### Learning Outcomes

- Use the elimination method with multiplication
- Express the solution of a dependent system of equations containing two variables

## Solve a system of equations when multiplication is necessary to eliminate a variable

Many times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.

[latex]\begin{array}{r}3x+4y=52\\5x+y=30\end{array}[/latex]

If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let’s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate the same variable in the other equation.

We do this with multiplication. Notice that the first equation contains the term [latex]4y[/latex], and the second equation contains the term [latex]y[/latex]. If you multiply the second equation by [latex]−4[/latex], when you add both equations the y variables will add up to [latex]0[/latex].

The following example takes you through all the steps to find a solution to this system.

### Example

Solve for [latex]x[/latex] and [latex]y[/latex].

**Equation A:** [latex]3x+4y=52[/latex]

**Equation B:** [latex]5x+y=30[/latex]

### Example

Solve the given system of equations by the **elimination method.**

[latex]\begin{array}{l}3x+5y=-11\hfill \\ x - 2y=11\hfill \end{array}[/latex]

Below is another video example of using the elimination method to solve a system of linear equations in which we multiply one of the equations be a constant.

It is worth demonstrating that there is more than one way to solve a system. Consider our first example. Instead of multiplying one equation in order to eliminate a variable when the equations were added, we could have multiplied both equations by different numbers.

Let’s remove the variable [latex]x[/latex] this time. Multiply Equation A by [latex]5[/latex] and Equation B by [latex]−3[/latex].

### Example

Solve for [latex]x[/latex] and [latex]y[/latex].

[latex]\begin{array}{r}3x+4y=52\\5x+y=30\end{array}[/latex]

These equations were multiplied by [latex]5[/latex] and [latex]−3[/latex] respectively, because that gave you terms that would add up to [latex]0[/latex]. Be sure to multiply all of the terms of the equation.

In the next example, we will see that sometimes we have to multiply both numbers by different numbers in order for one variable to be eliminated.

### Example

Solve the given system of equations in two variables by elimination.

Below is a summary of the general steps for using the elimination method to solve a system of equations.

### How To: Given a system of equations, solve using the elimination method

- Write both equations with
*x*and*y*-variables on the left side of the equal sign and constants on the right. - Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
- Solve the resulting equation for the remaining variable.
- Substitute that value into one of the original equations and solve for the second variable.
- Check the solution by substituting the values into the other equation.

### Try It

In the next example, we will show how to solve a system with fractions. As with single linear equations, the easiest way to solve is to clear the fractions first with the least common denominator.

### Example

Solve the given system of equations in two variables by elimination.

[latex]\begin{array}{l}\dfrac{x}{3}+\dfrac{y}{6}=3\hfill \\ \dfrac{x}{2}-\dfrac{y}{4}=\text{ }1\hfill \end{array}[/latex]

In the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions.

It is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as with the other methods we have learned for finding solutions to systems. Recall that a **dependent system** of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or elimination, the resulting equation will be an identity such as [latex]0=0[/latex]. The last example includes two equations that represent the same line and are therefore dependent.

### Example

Find a solution to the system of equations using the **elimination method**.

[latex]\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}[/latex]

In the following video, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a negative one first. Additionally, this system has an infinite number of solutions.

In our last video example, we present a system that is inconsistent; it has no solutions which means the lines the equations represent are parallel to each other.

### Try It

## Summary

Multiplication can be used to set up matching terms in equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation—not just the one term you are trying to eliminate.