Learning Objectives
By the end of this section, you will be able to:
- Solve an equation with constants on both sides
- Solve an equation with variables on both sides
- Solve an equation with variables and constants on both sides
Solve Equations with Constants on Both Sides
In all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we will learn to solve equations in which the variable terms, or constant terms, or both are on both sides of the equation.
Our strategy will involve choosing one side of the equation to be the “variable side”, and the other side of the equation to be the “constant side.” Then, we will use the Subtraction and Addition Properties of Equality to get all the variable terms together on one side of the equation and the constant terms together on the other side.
By doing this, we will transform the equation that began with variables and constants on both sides into the form \(ax=b\). We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.
Exercise \(\PageIndex{2}\)
Solve: \(3x+4=−8\).
- Answer
-
\(x=−4\)
Exercise \(\PageIndex{3}\)
Solve: \(5a+3=−37\).
- Answer
-
\(a=−8\)
Exercise \(\PageIndex{5}\)
Solve: \(5y−9=16\).
- Answer
-
\(y=5\)
Exercise \(\PageIndex{6}\)
Solve: \(3m−8=19\).
- Answer
-
\(m = 9\)
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.
Exercise \(\PageIndex{8}\)
Solve: \(6n=5n−10\).
- Answer
-
\(n = -10\)
Exercise \(\PageIndex{9}\)
Solve: \(-6c = -7c - 1\)
- Answer
-
\(c = -1\)
Exercise \(\PageIndex{10}\)
Solve: \(5y - 9 = 8y\)
- Answer
-
Exercise \(\PageIndex{11}\)
Solve: \(3p−14=5p\).
- Answer
-
\(p = -7\)
Exercise \(\PageIndex{12}\)
Solve: \(8m + 9 = 5m\)
- Answer
-
\(m = -3\)
Exercise \(\PageIndex{14}\)
Solve: \(12j = -4j + 32\)
- Answer
-
\(j = 2\)
Exercise \(\PageIndex{15}\)
Solve: \(8h = -4h + 12\)
- Answer
-
\(h = 1\)
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.
Exercise \(\PageIndex{16}\): How to Solve Equations with Variables and Constants on Both Sides
Solve: \(7x + 5 = 6x + 2\)
- Answer
-
Exercise \(\PageIndex{17}\)
Solve: \(12x+8=6x+2\).
- Answer
-
\(x=−1\)
Exercise \(\PageIndex{18}\)
Solve: \(9y+4=7y+12\).
- Answer
-
\(y=4\)
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.
Exercise \(\PageIndex{20}\)
Solve: \(8q - 5 = -4q + 7\)
- Answer
-
\(q = 1\)
Exercise \(\PageIndex{21}\)
Solve: \(7n - 3 = n + 3\)
- Answer
-
\(n = 1\)
Exercise \(\PageIndex{23}\)
Solve: \(2a - 2 = 6a + 18\)
- Answer
-
\(a = -5\)
Exercise \(\PageIndex{24}\)
Solve: \(4k -1 = 7k + 17\)
- Answer
-
\(k = -6\)
In the last example, we could have made the left side the “variable” side, but it would have led to a negative coefficient on the variable term. (Try it!) While we could work with the negative, there is less chance of errors when working with positives. The strategy outlined above helps avoid the negatives!
To solve an equation with fractions, we just follow the steps of our strategy to get the solution!
Exercise \(\PageIndex{25}\)
Solve: \(\frac{5}{4}x + 6 = \frac{1}{4}x - 2\)
- Answer
-
Since \(\frac{5}{4} > \frac{1}{4}\), make the left side the “variable” side and the right side the “constant” side.
Exercise \(\PageIndex{26}\)
Solve: \(\frac{7}{8}x - 12 = -\frac{1}{8}x - 2\)
- Answer
-
\(x = 10\)
Exercise \(\PageIndex{27}\)
Solve: \(\frac{7}{6}y + 11 = \frac{1}{6}y + 8\)
- Answer
-
\(y = -3\)
We will use the same strategy to find the solution for an equation with decimals.
Exercise \(\PageIndex{28}\)
Solve: \(7.8x+4=5.4x−8\).
- Answer
-
Since \(7.8>5.4\), make the left side the “variable” side and the right side the “constant” side.
Exercise \(\PageIndex{29}\)
Solve: \(2.8x + 12 = -1.4x - 9\)
- Answer
-
\(x = -5\)
Exercise \(\PageIndex{30}\)
Solve: \(3.6y + 8 = 1.2y - 4\)
- Answer
-
\(y = -5\)
Key Concepts
- Beginning Strategy for Solving an Equation 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.