2.3 Solve equations with variables and constants on both sides (Page 2/3)

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Solve: $6 n = 5 n - 10 .$

$n = −10$

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Solve: $−6 c = −7 c - 1 .$

$c = −1$

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Solve: $5 y - 9 = 8 y .$

Solution

The only constant is on the left and the $y$ ’s are on both sides. Let’s leave the constant on the left and get the variables to the right.


Subtract $5 y$ from both sides.
Simplify.
We have the y ’s on the right and the constants on the left. Divide both sides by 3.
Simplify.
Check:
Let $y = −3 .$

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Solve: $3 p - 14 = 5 p .$

$p = −7$

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Solve: $8 m + 9 = 5 m .$

$m = −3$

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Solve: $12 x = − x + 26 .$

Solution

The only constant is on the right, so let the left side be the “variable” side.


Remove the $− x$ from the right side by adding $x$ to both sides.
Simplify.
All the $x$ ’s are on the left and the constants are on the right. Divide both sides by 13.
Simplify.

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Solve: $12 j = −4 j + 32 .$

$j = 2$

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Solve: $8 h = −4 h + 12 .$

$h = 1$

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Solve equations with variables and constants on both sides

The next example will be the first to have variables and constants on both sides of the equation. It may take several steps to solve this equation, so we need a clear and organized strategy.

How to solve equations with variables and constants on both sides

Solve: $7 x + 5 = 6 x + 2 .$

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell on the left reads: “Step 1. Choose which side will the “variable” side—the other side will be the “constant” side.” The text in the second cell reads: “The variable terms are 7 x and 6 x. Since 7 is greater than 6, we will make the left side the “x” side and so the right side will be the “constant” side.” The third cell contains the equation 7 x plus 5 equals 6 x plus 2, and the left side of the equation is labeled “variable” written in red, and the right side of the equation is labeled “constant” written in red.

In the second row of the table, the first cell says: “Step 2. Collect the variable terms to the “variable” side of the equation, using the addition or subtraction property of equality.” In the second cell, the instructions say: “ With the right side as the “constant” side, the 6x is out of place, so subtract 6x from both sides. Combine like terms. Now the variable is only on the left side!” The third cell contains the original equation with 6x subtracted from both sides: 7 x minus 6 x plus 5 equals 6 x minus 6 x plus 2, with “minus 6 x” written in red on both sides. Below this is the same equation with like terms combined: x plus 5 equals 2.

In the third row of the table, the first cell says: “Step 3. Collect all the constants to the other side of the equation, using the addition or subtraction property of equality.” In the second cell, the instructions say: “The right side is the “constant” side, so the 5 is out of place. Subtract 5 from both sides. Simplify.” The third cell contains the equation x plus 5 minus 5 equals 2 minus 5, with “minus 5” written in red on both sides. Below this is the answer to the equation: x equals negative 3.

In the fourth row of the table, the first cell says: “Step 4. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.” In the second cell, the instructions say: “The coefficient of x is one. The equation is solved.” The third cell is blank.

In the fifth row of the table, the first cell says: “Step 5. Check.” The instructions in the second cell say: “Check. Let x equal negative 3. Simplify. Add.” In the third cell is the original equation again: 7 x plus 5 equals 6x plus 2. Below this is the same equation with negative 3 substituted in for x: 7 times negative 3 (in paretheses) plus 5 might equal 6 times negative 3 (in parentheses) plus 2, with the “times negative 3” written in red on both sides of the equation. Below this is the equation negative 21 plus 5 might equal negative 18 plus 2. On the last line is the equation negative 16 equals negative 16, with a check mark next to it.

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Solve: $12 x + 8 = 6 x + 2 .$

$x = −1$

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Solve: $9 y + 4 = 7 y + 12 .$

$y = 4$

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We’ll list the steps below so you can easily refer to them. But we’ll call this the ‘Beginning Strategy’ because we’ll be adding some steps later in this chapter.

Beginning strategy for solving equations with variables and constants on both sides of the equation.

Choose which side will be the “variable” side—the other side will be the “constant” side.
Collect the variable terms to the “variable” side of the equation, using the Addition or Subtraction Property of Equality.
Collect all the constants to the other side of the equation, using the Addition or Subtraction Property of Equality.
Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.
Check the solution by substituting it into the original equation.

In Step 1, a helpful approach is to make the “variable” side the side that has the variable with the larger coefficient. This usually makes the arithmetic easier.

Solve: $8 n - 4 = −2 n + 6 .$

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since $8 > −2$ , make the left side the “variable” side.
We don’t want variable terms on the right side—add $2 n$ to both sides to leave only constants on the right.
Combine like terms.
We don’t want any constants on the left side, so add $4$ to both sides.
Simplify.
The variable term is on the left and the constant term is on the right. To get the coefficient of $n$ to be one, divide both sides by 10.
Simplify.
Check:
Let $n = 1$ .

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Solve: $8 q - 5 = −4 q + 7 .$

$q = 1$

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Solve: $7 n - 3 = n + 3 .$

$n = 1$

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Solve: $7 a - 3 = 13 a + 7 .$

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since $13 > 7$ , make the right side the “variable” side and the left side the “constant” side.


Subtract $7 a$ from both sides to remove the variable term from the left.
Combine like terms.
Subtract $7$ from both sides to remove the constant from the right.
Simplify.
Divide both sides by $6$ to make $1$ the coefficient of $a$ .
Simplify.
Check:
Let $a = - \frac{5}{3}$ .

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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