How to simplify linear equations

How to simplify linear equations DEFAULT

2.4: Solving Linear Equations- Part II

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Learning Objectives

  • Solve general linear equations.
  • Identify and solve conditional equations, identities, and contradictions.
  • Clear decimals and fractions from equations.
  • Solve literal equations or formulas for a given variable.

Combining Like Terms and Simplifying

Linear equations typically are not given in standard form, so solving them requires additional steps. These additional steps include simplifying expressions on each side of the equal sign using the order of operations.

Same-Side Like Terms

We will often encounter linear equations where the expressions on each side of the equal sign can be simplified. Typically, this involves combining same-side like terms. If this is the case, then it is best to simplify each side first before solving.

Example \(\PageIndex{1}\)

Solve:

\(−4a+2−a=3−2\).

Solution:

First, combine the like terms on each side of the equal sign.

Answer:

The solution is \(\frac{1}{5}\).

Opposite-Side Like Terms

Given a linear equation in the form \(ax+b=cx+d\), we begin by combining like terms on opposite sides of the equal sign. To combine opposite-side like terms, use the addition or subtraction property of equality to effectively “move terms” from one side to the other so that they can be combined.

Example \(\PageIndex{2}\)

Solve:

\(−2y−3=5y+11\).

Solution:

To “move” the term \(5y\) to the left side, subtract it on both sides.

\(\begin{aligned} -2y-3&=5y+11 \\ -2y-3\color{Cerulean}{-5y}&=5y+11\color{Cerulean}{-5y} &\color{Cerulean}{Subtract\:5y\:from\:both\:sides.} \\ -7y-3&=11 \end{aligned}\)

From here, solve using the techniques developed previously.

Always check to see that the solution is correct by substituting the solution back into the original equation and simplifying to see if you obtain a true statement.

\(\begin{aligned} -2y-3&=5y+11 \\ -2(\color{OliveGreen}{-2}\color{black}{)-3}&=5(\color{OliveGreen}{-2}\color{black}{)+1} \\ 4-3&=-10+11 \\ 1&=1 \quad\color{Cerulean}{\checkmark} \end{aligned}\)

Answer:

The solution is \(-2\).

General Guidelines for Solving Linear Equations

When solving linear equations, the goal is to determine what value, if any, will produce a true statement when substituted in the original equation. Do this by isolating the variable using the following steps:

Step 1: Simplify both sides of the equation using the order of operations and combine all same-side like terms.

Step 2: Use the appropriate properties of equality to combine opposite-side like terms with the variable term on one side of the equation and the constant term on the other.

Step 3: Divide or multiply as needed to isolate the variable.

Step 4: Check to see if the answer solves the original equation.

Example \(\PageIndex{3}\)

Solve:

\(-\frac{1}{2}(10y-2)+3=14\)

Solution:

Simplify the linear expression on the left side before solving.

To check,

\(\begin{aligned} -\frac{1}{2}(10(\color{OliveGreen}{-2}\color{black}{)-2)+3}&=14 \\ -\frac{1}{2}(-20-2)+3&=14 \\ -\frac{1}{2}(-22)+3&=14 \\ 11+3&=14 \\ 14&=14\quad\color{Cerulean}{\checkmark} \end{aligned}\)

Answer:

The solution is \(-2\).

Example \(\PageIndex{4}\)

Solve:

\(5(3x+2)−2=−2(1−7x)\).

Solution:

First, simplify the expressions on both sides of the equal sign.

Answer:

The solution is \(−10\). The check is left as an exercise.

Exercise \(\PageIndex{1}\)

Solve:

\(6−3(4x−1)=4x−7\).

Answer

\(x=1\)

Conditional Equations, Identities, and Contradictions

There are three different types of equations. Up to this point, we have been solving conditional equations. These are equations that are true for particular values. An identity is an equation that is true for all possible values of the variable. For example,

\(x=x\quad\color{Cerulean}{Identity}\)

has a solution set consisting of all real numbers, \(R\). A contradiction is an equation that is never true and thus has no solutions. For example,

\(x+1=x\quad\color{Cerulean}{Contradiction}\)

has no solution. We use the empty set, \(∅\), to indicate that there are no solutions.

If the end result of solving an equation is a true statement, like \(0 = 0\), then the equation is an identity and any real number is a solution. If solving results in a false statement, like \(0 = 1\), then the equation is a contradiction and there is no solution.

Example \(\PageIndex{5}\)

Solve:

\(4(x+5)+6=2(2x+3)\).

Solution:

\(\begin{aligned} 4(x+5)+6&=2(2x+3)&\color{Cerulean}{Distribute.} \\ 4x\color{OliveGreen}{+20+6}&=4x+6&\color{Cerulean}{Combine\:same-side\:like\:terms.} \\ 4x+26&=4x+6 &\color{Cerulean}{Combine\:opposite-side\:like\:terms.} \\ 4x+26\color{Cerulean}{-4x}&=4x+6\color{Cerulean}{-4x} \\ 26&=6\quad\color{red}{x} &\color{Cerulean}{False} \end{aligned}\)

Answer:

\(∅\). Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.

Example \(\PageIndex{6}\)

Solve:

\(3(3y+5)+5=10(y+2)−y\).

Solution:

\(\begin{aligned} 3(3y+5)+5&=10(y+2)-y &\color{Cerulean}{Distribute.} \\ 9y\color{OliveGreen}{+15+5}&=10y+20-y &\color{Cerulean}{Combine\:same-side\:like\:terms.} \\ 9y+20&=9y+20 &\color{Cerulean}{Combine\:opposite-side\:like\:terms.} \\ 9y+20\color{Cerulean}{-9y}&=9y+20\color{Cerulean}{-9y} \\ 20&=20 \quad\color{Cerulean}{\checkmark} &\color{Cerulean}{True} \end{aligned}\)

Answer:

\(R\). Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.

If it is hard to believe that any real number is a solution to the equation in the previous example, then choose your favorite real number, and substitute it in the equation to see that it leads to a true statement. Choose \(x=7\) and check:

Exercise \(\PageIndex{2}\)

Solve:

\(−2(3x+1)−(x−3)=−7x+1\).

Answer

\(R\)

Clearing Decimals and Fractions

The coefficients of linear equations may be any real number, even decimals and fractions. When decimals and fractions are used, it is possible to use the multiplication property of equality to clear the coefficients in a single step. If given decimal coefficients, then multiply by an appropriate power of 10 to clear the decimals. If given fractional coefficients, then multiply both sides of the equation by the least common multiple of the denominators (LCD).

Example \(\PageIndex{7}\)

Solve:

\(2.3x+2.8=−1.2x+9.8\).

Solution:

Notice that all decimal coefficients are expressed with digits in the tenths place; this suggests that we can clear the decimals by multiplying both sides by \(10\). Take care to distribute \(10\) to each term on both sides of the equation.

\(\begin{aligned} \color{Cerulean}{10\cdot }\color{black}{(2.3x+2.8)} &=\color{Cerulean}{10\cdot}\color{black}{-1.2x+9.8} &\color{Cerulean}{Multiply\:both\:sides\:by\:10.} \\ \color{Cerulean}{10\cdot }\color{black}{2.3x+}\color{Cerulean}{10\cdot }\color{black}{2.8}&=\color{Cerulean}{10\cdot }\color{black}{(-1.2x)+}\color{Cerulean}{10\cdot}\color{black}{9.8} \\ 23x+28&=-12x+98 &\color{Cerulean}{Integer\:coefficients} \\ 23x+28\color{Cerulean}{+12x}&=-12x+98\color{Cerulean}{+12x} &\color{Cerulean}{Solve.} \\ 35x+28&=98 \\ 35x+28\color{Cerulean}{-28}&=98\color{Cerulean}{-28} \\ 35x&=70 \\ \frac{35x}{\color{Cerulean}{35}}&=\frac{70}{\color{Cerulean}{35}} \\ x&=2 \end{aligned}\)

Answer:

The solution is \(2\).

Example \(\PageIndex{8}\)

Solve:

\(\frac{1}{3}x+\frac{1}{5}=\frac{1}{5}x−1\).

Solution:

Clear the fractions by multiplying both sides by the least common multiple of the given denominators. In this case, the LCM\((3, 5)=15\).

Answer:

The solution is \(-9\).

It is important to know that these techniques only work for equations. Do not try to clear fractions when simplifying expressions. As a reminder

\(\begin{array}{c|c} {\underline{\color{Cerulean}{Expression}}}&{\underline{\color{Cerulean}{Equation}}} \\ {\frac{1}{2}x+\frac{5}{3}}&{\frac{1}{2}x+\frac{5}{3}=0} \end{array}\)

Solve equations and simplify expressions. If you multiply an expression by \(6\), you will change the problem. However, if you multiply both sides of an equation by 6, you obtain an equivalent equation.

\(\begin{array}{c|c} {\underline{\color{red}{Incorrect}}}&{\underline{\color{Cerulean}{Correct}}}\\{\frac{1}{2}x+\frac{5}{3}}&{\frac{1}{2}x+\frac{5}{3}=0} \\ {\neq\color{red}{6\cdot}\color{black}{\left( \frac{1}{2}x+\frac{5}{3} \right)}}&{\color{Cerulean}{6\cdot}\color{black}{\left( \frac{1}{2}x+\frac{5}{3} \right) =}\color{Cerulean}{6\cdot }\color{black}{0}} \\{=3x+10\quad\color{red}{x}}&{3x+10=10\quad\color{Cerulean}{\checkmark}} \end{array}\)

Literal Equations (Linear Formulas)

Algebra lets us solve whole classes of applications using literal equations, or formulas. Formulas often have more than one variable and describe, or model, a particular real-world problem. For example, the familiar formula \(D=rt\) describes the distance traveled in terms of the average rate and time; given any two of these quantities, we can determine the third. Using algebra, we can solve the equation for any one of the variables and derive two more formulas.

\(\begin{aligned} D&=rt \\ \frac{D}{\color{Cerulean}{r}}&=\frac{rt}{\color{Cerulean}{r}}&\color{Cerulean}{Divide\:both\:sides\:by\:r.} \\ \frac{D}{r}&=t \end{aligned}\)

If we divide both sides by \(r\), we obtain the formula \(t=Dr\). Use this formula to find the time, given the distance and the rate

\(\begin{aligned} D&=rt \\ \frac{D}{\color{Cerulean}{t}}&=\frac{rt}{\color{Cerulean}{t}}&\color{Cerulean}{Divide\:both\:sides\:by\:t.} \\ \frac{D}{t}&=r \end{aligned}\)

If we divide both sides by \(t\), we obtain the formula \(r=Dt\). Use this formula to find the rate, given the distance traveled and the time it takes to travel that distance. Using the techniques learned up to this point, we now have three equivalent formulas relating distance, average rate, and time:

\(D=rt\qquad t=\frac{D}{r}\qquad r=\frac{D}{t}\)

When given a literal equation, it is often necessary to solve for one of the variables in terms of the others. Use the properties of equality to isolate the indicated variable.

Example \(\PageIndex{9}\)

Solve for \(a\):

\(P=2a+b\).

Solution:

The goal is to isolate the variable \(a\).

\(\begin{aligned} P&=2a+b \\ P\color{Cerulean}{-b}&=2a+b\color{Cerulean}{-b} &\color{Cerulean}{Subtract\:b\:from\:both\:sides.} \\ P-b&=2a \\ \frac{P-b}{\color{Cerulean}{2}}&=\frac{2a}{\color{Cerulean}{2}} &\color{Cerulean}{Divide\:both\:sides\:by\:2.} \\ \frac{P-b}{2}&=a \end{aligned}\)

Answer:

\(a=\frac{P-b}{2}\)

Example \(\PageIndex{10}\)

Solve for \(y\):

\(z=\frac{x+y}{2}\).

Solution:

The goal is to isolate the variable \(y\).

\(\begin{aligned} z&=\frac{x+y}{2} \\ \color{Cerulean}{2\cdot}\color{black}{z}&=\color{Cerulean}{2\cdot}\color{black}{\frac{x+y}{2}}&\color{Cerulean}{Multiply\:both\:sides\:by\:2.} \\ 2z&=x+y \\ 2z\color{Cerulean}{-x}&=x+y\color{Cerulean}{-x}&\color{Cerulean}{Subtract\:x\:from\:both\:sides.} \\ 2z-x&=y \end{aligned}\)

Answer:

\(y=2z-x\)

Exercise \(\PageIndex{3}\)

Solve for \(b\):

\(2a−3b=c\).

Answer

\(b=\frac{2a−c}{3}\)

Key Takeaways

  • Solving general linear equations involves isolating the variable, with coefficient \(1\), on one side of the equal sign.
  • The steps for solving linear equations are:
    • Simplify both sides of the equation and combine all same-side like terms.
    • Combine opposite-side like terms to obtain the variable term on one side of the equal sign and the constant term on the other.
    • Divide or multiply as needed to isolate the variable.
    • Check the answer.
  • Most linear equations that you will encounter are conditional and have one solution.
  • If solving a linear equation leads to a true statement like \(0 = 0\), then the equation is an identity and the solution set consists of all real numbers, \(R\).
  • If solving a linear equation leads to a false statement like \(0 = 5\), then the equation is a contradiction and there is no solution, \(∅\).
  • Clear fractions by multiplying both sides of a linear equation by the least common multiple of all the denominators. Distribute and multiply all terms by the LCD to obtain an equivalent equation with integer coefficients.
  • Given a formula, solve for any variable using the same techniques for solving linear equations. This works because variables are simply representations of real numbers.

Exercise \(\PageIndex{4}\) Checking for Solutions

Is the given value a solution to the linear equation?

  1. \(2(3x+5)−6=3x−8; x=−4 \)
  2. \(−x+17−8x=9−x; x=−1 \)
  3. \(4(3x−7)−3(x+2)=−1; x=\frac{1}{3}\)
  4. \(−5−2(x−5)=−(x+3); x=−8 \)
  5. \(7−2(\frac{1}{2}x−6)=x−1; x=10 \)
  6. \(3x−\frac{2}{3}(9x−2)=0; x=\frac{4}{9}\)
Answer

1. Yes

3. No

5. Yes

Exercise \(\PageIndex{5}\) Solving Linear Equations

Solve.

  1. \(4x−7=7x+5\)
  2. \(−5x+3=−8x−9\)
  3. \(3x−5=2x−17\)
  4. \(−2y−52=3y+13\)
  5. \(−4x+2=7x−20\)
  6. \(4x−3=6x−15\)
  7. \(9x−25=12x−25\)
  8. \(12y+15=−6y+23\)
  9. \(1.2x−0.7=3x+4.7\)
  10. \(2.1x+6.1=−1.3x+4.4\)
  11. \(2.02x+4.8=14.782−1.2x\)
  12. \(−3.6x+5.5+8.2x=6.5+4.6x\)
  13. \(\frac{1}{2}x−\frac{2}{3}=x+\frac{1}{5}\)
  14. \(\frac{1}{3}x−\frac{1}{2}=−\frac{1}{4}x−\frac{1}{3}\)
  15. \(−\frac{1}{10}y+\frac{2}{5}=\frac{1}{5}y+\frac{3}{10}\)
  16. \(x−\frac{20}{3}=\frac{5}{2}x+\frac{5}{6}\)
  17. \(\frac{2}{3}y+\frac{1}{2}=\frac{5}{8}y+\frac{37}{24}\)
  18. \(\frac{1}{3}+\frac{4}{3}x=\frac{10}{7}x+\frac{1}{3}−\frac{2}{21}x\)
  19. \(\frac{8}{9}−\frac{11}{18}x=\frac{7}{6}−12x\)
  20. \(\frac{1}{3}−9x=\frac{4}{9}+\frac{1}{2}x\)
  21. \(12x−5+9x=44\)
  22. \(10−6x−13=12\)
  23. \(−2+4x+9=7x+8−2x\)
  24. \(20x−5+12x=6−x+7\)
  25. \(3a+5−a=2a+7\)
  26. \(−7b+3=2−5b+1−2b\)
  27. \(7x−2+3x=4+2x−2\)
  28. \(−3x+8−4x+2=10\)
  29. \(6x+2−3x=−2x−13\)
  30. \(3x−0.75+0.21x=1.24x+7.13\)
  31. \(−x−2+4x=5+3x−7\)
  32. \(−2y−5=8y−6−10y\)
  33. \(\frac{1}{10}x−\frac{1}{3}=\frac{1}{30}−\frac{1}{15}x−\frac{7}{15}\)
  34. \(\frac{5}{8}−\frac{4}{3}x+\frac{1}{3}=−\frac{3}{9}x−\frac{1}{4}+\frac{1}{3}x\)
Answer

1. \(−4\)

3. \(−12\)

5. \(2\)

7. \(0\)

9. \(−3\)

11. \(3.1\)

13. \(−\frac{26}{15}\)

15. \(\frac{1}{3}\)

17. \(25\)

19. \(−\frac{5}{2}\)

21. \(\frac{7}{3}\)

23. \(−1\)

25. \(∅\)

27. \(\frac{1}{2}\)

29. \(−3\)

31. \(R\)

33. \(−\frac{3}{5}\)

Exercise \(\PageIndex{6}\) Solving Linear Equations Involving Parentheses

Solve.

  1. \(−5(2y−3)+2=12\)
  2. \(3(5x+4)+5x=−8\)
  3. \(4−2(x−5)=−2\)
  4. \(10−5(3x+1)=5(x−4)\)
  5. \(9−(x+7)=2(x−1)\)
  6. \(−5(2x−1)+3=−12\)
  7. \(3x−2(x+1)=x+5\)
  8. \(5x−3(2x−1)=2(x−3)\)
  9. \(−6(x−1)−3x=3(x+8)\)
  10. \(−\frac{3}{5}(5x+10)=\frac{1}{2}(4x−12)\)
  11. \(3.1(2x−3)+0.5=22.2\)
  12. \(4.22−3.13(x−1)=5.2(2x+1)−11.38\)
  13. \(6(x−2)−(7x−12)=14\)
  14. \(−9(x−3)−3x=−3(4x+9)\)
  15. \(3−2(x+4)=−3(4x−5)\)
  16. \(12−2(2x+1)=4(x−1)\)
  17. \(3(x+5)−2(2x+3)=7x+9\)
  18. \(3(2x−1)−4(3x−2)=−5x+10\)
  19. \(−3(2a−3)+2=3(a+7)\)
  20. \(−2(5x−3)−1=5(−2x+1)\)
  21. \(\frac{1}{2}(2x+1)−\frac{1}{4}(8x+2)=3(x−4)\)
  22. \(−\frac{2}{3}(6x−3)−\frac{1}{2}=\frac{3}{2}(4x+1)\)
  23. \(\frac{1}{2}(3x−1)+\frac{1}{3}(2x−5)=0\)
  24. \(\frac{1}{3}(x−2)+\frac{1}{5}=\frac{1}{9}(3x+3)\)
  25. \(−2(2x−7)−(x+3)=6(x−1)\)
  26. \(10(3x+5)−5(4x+2)=2(5x+20)\)
  27. \(2(x−3)−6(2x+1)=−5(2x−4)\)
  28. \(5(x−2)−(4x−1)=−2(3−x)\)
  29. \(6(3x−2)−(12x−1)+4=0\)
  30. \(−3(4x−2)−(9x+3)−6x=0\)
Answer

1. \(\frac{1}{2}\)

3. \(8\)

5. \(\frac{4}{3}\)

7. \(∅\)

9. \(−\frac{3}{2}\)

11. \(5\)

13. \(−14\)

15. \(2\)

17. \(0\)

19. \(−\frac{10}{9}\)

21. \(3\)

23. \(1\)

25. \(\frac{17}{11}\)

27. \(∅\)

29. \(\frac{7}{6}\)

Exercise \(\PageIndex{7}\) Literal Equations

Solve for the indicated variable.

  1. Solve for \(w\): \(A=l⋅w\).
  2. Solve for \(a\): \(F=ma\).
  3. Solve for \(w\): \(P=2l+2w\).
  4. Solve for \(r\): \(C=2πr\).
  5. Solve for \(b\): \(P=a+b+c\).
  6. Solve for \(C\): \(F=\frac{9}{5}C+32\).
  7. Solve for \(h\): \(A=\frac{1}{2}bh\).
  8. Solve for \(t\): \(I=Prt\).
  9. Solve for \(y\): \(ax+by=c\).
  10. Solve for \(h\): \(S=2πr^{2}+2πrh\).
  11. Solve for \(x\): \(z=\frac{2x+y}{5}\).
  12. Solve for \(c\): \(a=3b−\frac{2c}{3}\).
  13. Solve for \(b\): \(y=mx+b\).
  14. Solve for \(m\): \(y=mx+b\).
  15. Solve for \(y\): \(3x−2y=6\).
  16. Solve for \(y\): \(−5x+2y=12\).
  17. Solve for \(y\): \(\frac{x}{3}−\frac{y}{5}=1\).
  18. Solve for \(y\): \(\frac{3}{4}x−\frac{1}{5}y=\frac{1}{2}\).
Answer

1. \(w=\frac{A}{l}\)

3. \(w=\frac{P−2l}{2}\)

5. \(b=P−a−c \)

7. \(h=\frac{2A}{b}\)

9. \(y=\frac{−ax+c}{b}\)

11. \(x=\frac{5z−y}{2}\)

13. \(b=y−mx\)

15. \(y=\frac{3x−6}{2}\)

17. \(y=\frac{5x−15}{3}\)

Exercise \(\PageIndex{8}\) Literal Equations

Translate the following sentences into linear equations and then solve.

  1. The sum of \(3x\) and \(5\) is equal to the sum of \(2x\) and \(7\).
  2. The sum of \(−5x\) and \(6\) is equal to the difference of \(4x\) and \(2\).
  3. The difference of \(5x\) and \(25\) is equal to the difference of \(3x\) and \(51\).
  4. The sum of \(\frac{1}{2}x\) and \(\frac{3}{4}\) is equal to \(\frac{2}{3}x\).
  5. A number \(n\) divided by \(5\) is equal to the sum of twice the number and \(3\).
  6. Negative ten times a number \(n\) is equal to the sum of three times the number and \(13\).
Answer

1. \(3x+5=2x+7\); \(x=2\)

3. \(5x−25=3x−51\); \(x=−13\)

5. \(\frac{n}{5}=2n+3\); \(n=−\frac{5}{3}\)

Exercise \(\PageIndex{9}\) Discussion Board Topics

  1. What is the origin of the word algebra?
  2. What is regarded as the main business of algebra?
  3. Why is solving equations such an important algebra topic?
  4. Post some real-world linear formulas not presented in this section.
  5. Research and discuss the contributions of Diophantus of Alexandria.
  6. Create an identity or contradiction of your own and share on the discussion board. Provide a solution and explain how you found it.
Answer

1. Answers may vary

3. Answers may vary

5. Answers may vary

Sours: https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Beginning_Algebra_(Redden)/02%3A_Linear_Equations_and_Inequalities/2.04%3A_Solving_Linear_Equations-_Part_II
Solve each of the following equations.
Show Discussion

In the following problems we will describe in detail the first problem and the leave most of the explanation out of the following problems.


a\(3\left( {x + 5} \right) = 2\left( { - 6 - x} \right) - 2x\) Show Solution

For this problem there are no fractions so we don’t need to worry about the first step in the process. The next step tells to simplify both sides. So, we will clear out any parenthesis by multiplying the numbers through and then combine like terms.

\[\begin{align*}3\left( {x + 5} \right) & = 2\left( { - 6 - x} \right) - 2x\\ 3x + 15 & = - 12 - 2x - 2x\\ 3x + 15 & = - 12 - 4x\end{align*}\]

The next step is to get all the \(x\)’s on one side and all the numbers on the other side. Which side the \(x\)’s go on is up to you and will probably vary with the problem. As a rule of thumb, we will usually put the variables on the side that will give a positive coefficient. This is done simply because it is often easy to lose track of the minus sign on the coefficient and so if we make sure it is positive we won’t need to worry about it.

So, for our case this will mean adding 4\(x\) to both sides and subtracting 15 from both sides. Note as well that while we will actually put those operations in this time we normally do these operations in our head.

\[\begin{align*}\require{color} 3x + 15 & = - 12 - 4x\\ 3x + 15 {\color{Red} - 15}{\color{Blue} + 4x} & = - 12 - 4x {\color{Blue} + 4x} {\color{Red} - 15}\\ 7x & = - 27\end{align*}\]

The next step says to get a coefficient of 1 in front of the \(x\). In this case we can do this by dividing both sides by a 7.

\[\begin{align*}\frac{{7x}}{7} &= \frac{{ - 27}}{7}\\ x & = - \frac{{27}}{7}\end{align*}\]

Now, if we’ve done all of our work correct \(x = - \frac{{27}}{7}\) is the solution to the equation.

The last and final step is to then check the solution. As pointed out in the process outline we need to check the solution in the original equation. This is important, because we may have made a mistake in the very first step and if we did and then checked the answer in the results from that step it may seem to indicate that the solution is correct when the reality will be that we don’t have the correct answer because of the mistake that we originally made.

The problem of course is that, with this solution, that checking might be a little messy. Let’s do it anyway.

\[\begin{align*}3\left( { - \frac{{27}}{7} + 5} \right) & \mathop = \limits^? 2\left( { - 6 - \left( { - \frac{{27}}{7}} \right)} \right) - 2\left( { - \frac{{27}}{7}} \right)\\ 3\left( {\frac{8}{7}} \right) & \mathop = \limits^? 2\left( { - \frac{{15}}{7}} \right) + \frac{{54}}{7}\\ \frac{{24}}{7} & = \frac{{24}}{7}\hspace{0.5in}{\mbox{OK}}\end{align*}\]

So, we did our work correctly and the solution to the equation is,

\[x = - \frac{{27}}{7}\]

Note that we didn’t use the solution set notation here. For single solutions we will rarely do that in this class. However, if we had wanted to the solution set notation for this problem would be,

\[\left\{ { - \frac{{27}}{7}} \right\}\]

Before proceeding to the next problem let’s first make a quick comment about the “messiness’ of this answer. Do NOT expect all answers to be nice simple integers. While we do try to keep most answer simple often they won’t be so do NOT get so locked into the idea that an answer must be a simple integer that you immediately assume that you’ve made a mistake because of the “messiness” of the answer.


b\(\displaystyle \frac{{m - 2}}{3} + 1 = \frac{{2m}}{7}\) Show Solution

Okay, with this one we won’t be putting quite as much explanation into the problem.

In this case we have fractions so to make our life easier we will multiply both sides by the LCD, which is 21 in this case. After doing that the problem will be very similar to the previous problem. Note as well that the denominators are only numbers and so we won’t need to worry about division by zero issues.

Let’s first multiply both sides by the LCD.

\[\begin{align*}21\left( {\frac{{m - 2}}{3} + 1} \right) & = \left( {\frac{{2m}}{7}} \right)21\\ 21\left( {\frac{{m - 2}}{3}} \right) + 21\left( 1 \right) & = \left( {\frac{{2m}}{7}} \right)21\\ 7\left( {m - 2} \right) + 21 & = \left( {2m} \right)\left( 3 \right)\end{align*}\]

Be careful to correctly distribute the 21 through the parenthesis on the left side. Everything inside the parenthesis needs to be multiplied by the 21 before we simplify. At this point we’ve got a problem that is similar the previous problem and we won’t bother with all the explanation this time.

\[\begin{align*}7\left( {m - 2} \right) + 21 & = \left( {2m} \right)\left( 3 \right)\\ 7m - 14 + 21 & = 6m\\ 7m + 7 & = 6m\\ m & = - 7\end{align*}\]

So, it looks like \(m = - 7\) is the solution. Let’s verify it to make sure.

\[\begin{align*}\frac{{ - 7 - 2}}{3} + 1 & \mathop = \limits^? \frac{{2\left( { - 7} \right)}}{7}\\ \frac{{ - 9}}{3} + 1 & \mathop = \limits^? - \frac{{14}}{7}\\ - 3 + 1 &\mathop = \limits^? - 2\\ - 2 & = - 2\hspace{0.5in}{\mbox{OK}}\end{align*}\]

So, it is the solution.


c\(\displaystyle \frac{5}{{2y - 6}} = \frac{{10 - y}}{{{y^2} - 6y + 9}}\) Show Solution

This one is similar to the previous one except now we’ve got variables in the denominator. So, to get the LCD we’ll first need to completely factor the denominators of each rational expression.

\[\frac{5}{{2\left( {y - 3} \right)}} = \frac{{10 - y}}{{{{\left( {y - 3} \right)}^2}}}\]

So, it looks like the LCD is \(2{\left( {y - 3} \right)^2}\). Also note that we will need to avoid \(y = 3\) since if we plugged that into the equation we would get division by zero.

Now, outside of the \(y\)’s in the denominator this problem works identical to the previous one so let’s do the work.

\[\begin{align*}\left( 2 \right){\left( {y - 3} \right)^2}\left( {\frac{5}{{2\left( {y - 3} \right)}}} \right) & = \left( {\frac{{10 - y}}{{{{\left( {y - 3} \right)}^2}}}} \right)\left( 2 \right){\left( {y - 3} \right)^2}\\ 5\left( {y - 3} \right) & = 2\left( {10 - y} \right)\\ 5y - 15 & = 20 - 2y\\ 7y & = 35\\ & y = 5\end{align*}\]

Now the solution is not \(y = 3\) so we won’t get division by zero with the solution which is a good thing. Finally, let’s do a quick verification.

\[\begin{align*}\frac{5}{{2\left( 5 \right) - 6}} & \mathop = \limits^? \frac{{10 - 5}}{{{5^2} - 6\left( 5 \right) + 9}}\\ \frac{5}{4} & = \frac{5}{4}\hspace{0.5in}{\mbox{OK}}\end{align*}\]

d\(\displaystyle \frac{{2z}}{{z + 3}} = \frac{3}{{z - 10}} + 2\) Show Solution

In this case it looks like the LCD is \(\left( {z + 3} \right)\left( {z - 10} \right)\) and it also looks like we will need to avoid \(z = - 3\) and \(z = 10\) to make sure that we don’t get division by zero.

Let’s get started on the work for this problem.

\[\begin{align*}\left( {z + 3} \right)\left( {z - 10} \right)\left( {\frac{{2z}}{{z + 3}}} \right) & = \left( {\frac{3}{{z - 10}} + 2} \right)\left( {z + 3} \right)\left( {z - 10} \right)\\ 2z\left( {z - 10} \right) & = 3\left( {z + 3} \right) + 2\left( {z + 3} \right)\left( {z - 10} \right)\\ 2{z^2} - 20z & = 3z + 9 + 2\left( {{z^2} - 7z - 30} \right)\end{align*}\]

At this point let’s pause and acknowledge that we’ve got a z2 in the work here. Do not get excited about that. Sometimes these will show up temporarily in these problems. You should only worry about it if it is still there after we finish the simplification work.

So, let’s finish the problem.

\[\begin{align*}\require{cancel} \cancel{{2{z^2}}} - 20z & = 3z + 9 + \cancel{{2{z^2}}} - 14z - 60\\ - 20z & = - 11z - 51\\ 51 & = 9z\\ \frac{{51}}{9} & = z\\ & \frac{{17}}{3} = z\end{align*}\]

Notice that the z2 did in fact cancel out. Now, if we did our work correctly \(z = \frac{{17}}{3}\) should be the solution since it is not either of the two values that will give division by zero. Let’s verify this.

\[\begin{align*}\frac{{2\left( {\frac{{17}}{3}} \right)}}{{\frac{{17}}{3} + 3}} & \mathop = \limits^? \frac{3}{{\frac{{17}}{3} - 10}} + 2\\ \frac{{\,\,\frac{{34}}{3}\,\,}}{{\frac{{26}}{3}}} & \mathop = \limits^? \frac{3}{{ - \frac{{13}}{3}}} + 2\\ \frac{{34}}{3}\left( {\frac{3}{{26}}} \right) & \mathop = \limits^? 3\left( { - \frac{3}{{13}}} \right) + 2\\ \frac{{17}}{{13}} & = \frac{{17}}{{13}}\hspace{0.5in}{\mbox{OK}}\end{align*}\]

The checking can be a little messy at times, but it does mean that we KNOW the solution is correct.

Sours: https://tutorial.math.lamar.edu/classes/alg/solvelineareqns.aspx
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Solving One-Step Linear Equations: Adding & Subtracting

Purplemath

"Linear" equations are equations with just a plain old variable like "x", rather than something more complicated like x2, or x/y, or square roots, or other more-complicated expressions. Linear equations are the simplest equations that you'll deal with.

You've probably already solved linear equations; you just didn't know it. Back in your early years, when you were learning addition, your teacher probably gave you worksheets to complete that had exercises like the following:

Fill in the box: □ + 3 = 5

Fill in the box: □ + 3 = 5

Once you'd learned your addition facts well enough, you knew that you had to put a "2" inside the box.

Solving equations works in much the same way, but now we have to figure out what goes into the x, instead of what goes into the box. However, since we're older now than when we were filling in boxes, the equations can also be much more complicated, and therefore the methods we'll use to solve the equations will be a bit more advanced.

In general, to solve an equation for a given variable, we need to "undo" whatever has been done to the variable. We do this in order to get the variable by itself; in technical terms, we are "isolating" the variable. This results in the equation being rearranged to say "(variable) equals (some number)", where (some number) is the answer they're looking for. For instance:

The variable is the letter x. To solve this equation, I need to get the x by itself; that is, I need to get x on one side of the "equals" sign, and some number on the other side.

Since I want just x on the one side, this means that I don't like the "plus six" that's currently on the same side as the x. Since the 6 is added to the x, I need to subtract this 6 to get rid of it. That is, I will need to subtract a 6 from the x in order to "undo" their having added a 6 to it.

This brings up the most important consideration with equations:

No matter what kind of equation we're dealing with — linear or otherwise — whatever we do to the one side of the equation, we must do the exact same thing to the other side of the equation. Equations are like toddlers in this respect:

We have to be totally, totally fair to the two sides, or unhappiness will ensue!

Whatever you do to an equation, do the EXACT SAME thing to BOTH sides of that equation!

Probably the best way to keep track of this subtraction of the 6 from both sides is to format your work this way:

animation: solving x + 6 = –3 by subtracting 6 from either side; this can be thought of as writing a new line showing x + 6 – 6 = –3 – 6, and then simplifying to get x = –9

The above image is animated on the "live" page.

What you see here is that I've subtracted 6 from both sides, drawn a horizontal "equals" bar underneath the entire equation, and then added down. On the left-hand side (LHS) of the equation, this gives me:

x plus nothing is x, and 6 minus 6 is zero

On the right-hand side (RHS) of the equation, I have:

–3 plus –6 is –9

The solution is the last line of my work; namely:


The same "undo" procedure works for equations in which the variable has been paired with a subtraction.

  • Solve x – 3 = –5

The variable is on the left-hand side (LHS) of the equation, and it's paired with a "subtract three". Since I want to get x by itself, I don't like the "3" that's currently subtracted from it. The opposite of subtraction is addition, so I'll undo the "subtract 3" by adding 3 to both sides of the equation, and then adding down to simplify to get my answer:

original equation: x – 3 = – 5; add 3 to either side of the equation: x – 3 + 3 = –5 + 3; add down to get x = –2

Then my answer is:


You may be instructed to "check your solutions", at least in the early stages of learning how to solve equations. To do this "checking", you need only plug your answer into the original equation, and make sure that you end up with a true statement. (This is, after all, the definition of the solution to an equation; namely, the solution is any value, or set of values [for more complicated equations, later on], which makes the original equation a true statement.)

So, to check my solution to the above equation, you'd plug "–2" in place of the x in left-hand side (LHS) of the original equation, and check that this simplifies to give the original value for the right-hand side (RHS) of the equation:

Checking:

LHS: (–2) – 3 = –5

RHS: –5

Because each side of the original equation now evaluates to the exact same thing, this confirms that the solution is indeed correct.


  • Solve 4 = x – 3, and check your solution.

This time, the variable is on the right-hand side (RHS) of the equation. That's okay; it doesn't matter where the variable is, as long as I get isolate it (that is, as long as I can get it by itself on one side of the "equals" sign).

In this equation, I've got a three that's subtracted from the variable. To undo the subtraction, I'll add three to either side of the equation.

4 = x - 3
+3 + 3
----------
7 = x

(I could have written the right-hand side, after adding down, as "x + 0", but "plus zero" is customarily ignored. That's why I carried down only the x on the right-hand side.)

Now, as part of my hand-in work, I need to show that I've checked this solution by plugging it back into the RHS of the original equation, and confirming that I end up with the LHS of the original equation; that is, that I end up with 4:

Checking:

RHS: (7) – 3 = 4 = LHS

The "checking" part is what I just did above. I've made sure to label things clearly, so the grader is able to find my "check" (so I'll get full credit on the exercise). My final answer is:


When I solved the last exercise above, the variable had ended up on the right-hand side of the "equals" sign. But in my solution, I wrote the answer with the variable on the left-hand side of the "equals" sign. This is pretty standard. When you're solving, the variable will end up wherever it ends up. When you're writing out the solution, the variable goes on the left. Why? Because.


This equation is almost solved. But not quite. I don't have plain old x on the right-hand side; instead, I've got –x. What to do?

I can kind-of think of the –x as being 0 – x. So what would happen if I added x to each side of the equation?

2 = –x
+x +x
-------
x + 2 = 0

Okay; that helped. By taking the variable and "adding it over to the other side", I've now got the variable in a format I like. And this has also converted the original equation into a simple one-step equation. I'll get rid of the 2 from the left-hand side by "subtracting it over to" the right-hand side:

x + 2 = 0
-2 = -2
----------
x = -2

This answer makes sense. If the negative of the variable equalled a positive two, then the positive of the variable should equal a negative two. So my answer is:


Technically, that last example was a two-step equation, because solving it required adding one thing to both sides of the equation, and then subtracting another thing to both sides. The important thing to notice is that you can add and subtract variables to the other side of an equation, just like you can add and subtract numbers to the other side. The exact same methods work with both variables and numbers.


You can use the Mathway widget below to practice solving a linear equation by adding or subtracting. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)

(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)



URL: https://www.purplemath.com/modules/solvelin.htm

Sours: https://www.purplemath.com/modules/solvelin.htm
Linear Equations - Algebra

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Simplify equations linear to how

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Art of Problem Solving: Simplifying Linear Expressions

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Similar news:

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