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# Linear Systems with Addition or Subtraction

## Solve systems using elimination of one variable

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Practice Linear Systems with Addition or Subtraction
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Solving Systems Without Multiplying

Three times a number plus 5 equals twice another number. \begin{align*}-4\end{align*} times the same number minus 2 equals \begin{align*}-2\end{align*} times the other number. What are the two numbers?

### Solving Systems Without Multiplying

In this lesson we will be looking at systems in which the two equations contain coefficients of one variable that are additive inverses (opposites) of one another.

#### Solve the following systems

Solve the system using Linear Combination:

\begin{align*}2x-3y &= -9\\ 5x +3y &= 30\end{align*}

Notice that the coefficients of the \begin{align*}y\end{align*} terms are opposites. When we add the two equations together, these terms will be eliminated because their sum is \begin{align*}0y = 0\end{align*}.

\begin{align*}& \quad \ \ 2x- \cancel{3y} = -9\\ & \ + \underline{5x + \cancel{3y} = 30}\\ & \qquad \quad \ \ 7x = 21\end{align*}

Now we can solve for \begin{align*}x\end{align*}:

\begin{align*}7x &= 21\\ x &= 3\end{align*}

Now that we have found \begin{align*}x\end{align*}, we can plug this value into either equation to find \begin{align*}y\end{align*}:

\begin{align*}2(3)-3y &= -9 \qquad \qquad \quad 5(3)+3y =30\\ 6-3y &= -9 \qquad \qquad \quad \quad 15+3y =30\\ -3y &= -15 \qquad or \qquad \qquad \ \ 3y = 15\\ y &=5 \qquad \qquad \qquad \qquad \quad \ y = 5 \end{align*}

The solution is therefore: (3, 5).

Remember to check your answer:

\begin{align*}2(3)-3(5) &= 6-15 = -9\\ 5(3)+3(5) &= 15+15= 30 \end{align*}

Solve the system using Linear Combination:

\begin{align*}x+4y &= 2\\ -x-5y &= -3\end{align*}

Notice that the coefficients of the \begin{align*}x\end{align*} terms are opposites. When we add the two equations together, these terms will be eliminated because their sum is \begin{align*}0x=0\end{align*}.

\begin{align*}& \quad \ \ \cancel{x}+4y = 2\\ & + \underline{- \cancel{x} - 5y = -3}\\ & \qquad \quad -y = -1\end{align*}

Now we can solve for \begin{align*}y\end{align*}:

\begin{align*}-y &= -1\\ y &= 1\end{align*}

Now that we have found \begin{align*}y\end{align*}, we can plug this value into either equation to find \begin{align*}x\end{align*}:

\begin{align*}& x+4(1) = 2 \qquad \quad \ \ \ -x-5(1) = -3\\ & \quad \ x+4 = 2 \qquad \qquad \quad \ -x-5 = -3\\ & \quad \quad \quad x = -2 \qquad or \qquad \quad \ -x=2\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x = -2\end{align*}

The solution is therefore: (-2, 1).

Remember to check your answer:

\begin{align*}-2+4(1) &= -2+4 =2\\ -(-2)-5(1) &= 2-5 =-3 \end{align*}

Solve the system using Linear Combination:

\begin{align*}2x+y &= 2\\ -3x+y &= -18\end{align*}

In this case the coefficients of the \begin{align*}y\end{align*} terms are the same, not opposites. One way to solve this system using linear combination would be to subtract the second equation from the first instead of adding it. Sometimes subtraction results in more errors, however, particularly when negative numbers are involved. Instead of subtracting, multiply the second equation by -1 and then add them together.

\begin{align*}-1(-3x+y &= - 18)\\ 3x-y &= 18\end{align*}

Essentially, we changed all of the signs of the terms in this equation.

Now we can add the two equations together to eliminate \begin{align*}y\end{align*}:

\begin{align*}& \quad \ 2x + \cancel{y} = 2\\ & \ + \underline{3x-\cancel{y} = 18}\\ & \qquad \quad 5x=20\end{align*}

Now we can solve for \begin{align*}x\end{align*}:

\begin{align*}5x &= 20\\ x &= 4\end{align*}

Now that we have found \begin{align*}x\end{align*}, plug this value into either equation to find \begin{align*}y\end{align*}:

\begin{align*}2(4) + y &= 2 \qquad \quad -3(4)+y = -18\\ 8+y &= 2 \quad \ \ or \quad \ -12+y = -18\\ y &= -6 \qquad \qquad \quad \ \quad \ y=-6\end{align*}

The solution is therefore: (4, -6).

Remember to check your answer:

\begin{align*}2(4)+(-6) &= 8-6 = 2\\ -3(4) + (-6) &= -12 -6 = -18 \end{align*}

### Examples

#### Example 1

Earlier, you were asked what were the other two numbers.

The system of linear equations represented by this situation is:

\begin{align*}3x + 5 &= 2y\\ -4x - 2 = -2y\end{align*}

If we add these two equations we get:

\begin{align*}-x + 3 = 0\end{align*} or \begin{align*}x = 3\end{align*}

We can now substitute \begin{align*}x = 3\end{align*} into either of the original equations to get \begin{align*}y = 7\end{align*}.

Solve the following systems using Linear Combinations.

#### Example 2

\begin{align*}4x+5y &= 8\\ -2x-5y &= 6\end{align*}

First we can add the two equations together to eliminate \begin{align*}y\end{align*} and solve for \begin{align*}x\end{align*}:

\begin{align*}& \qquad \qquad 4x+\cancel{5y}= 8\\ & \qquad \ + \underline{-2x-\cancel{5y} = 6}\\ & \qquad \qquad \quad \quad \ 2x= 14\\ & \qquad \qquad \qquad \ \ x=7\end{align*}

Substitute \begin{align*}x\end{align*} into one equation to find \begin{align*}y\end{align*}:

\begin{align*}4(7)+5y &= 8\\ 28 + 5y &= 8\\ 5y &= -20\\ y &= -4\end{align*}

Solution: (7, -4)

#### Example 3

\begin{align*}2x+3y &= 3\\ 2x-y &= 23\end{align*}

This time we need to begin by multiplying the second equation by -1 to get \begin{align*}-2x+y=-23\end{align*}. Now we can add the two equations together to eliminate \begin{align*}x\end{align*} and solve for \begin{align*}y\end{align*}:

\begin{align*}& \qquad \ \cancel{2x}+3y= 3\\ & \quad + \underline{- \cancel{2x}+y = -23}\\ & \qquad \qquad \ \ 4y= -20\\ & \qquad \qquad \quad y=-5\end{align*}

Substitute \begin{align*}y\end{align*} into one equation to find \begin{align*}x\end{align*}:

\begin{align*}2x+(3-5) &= 3\\ 2x-15 &= 3\\ 2x &= 18\\ x &= 9\end{align*}

Solution: (9, -5)

#### Example 4

\begin{align*}2x+3y &= -6\\ y &= 2x-2\end{align*}

In this example, the second equation is not written in standard form. We must first rewrite this equation in standard form so that the variable will align vertically when we add the equations together. The second equation should be \begin{align*}-2x+y=-2\end{align*} after we subtract \begin{align*}2x\end{align*} from both sides. Now we can add the two equations together to eliminate \begin{align*}x\end{align*} and solve for \begin{align*}y\end{align*}:

\begin{align*}& \quad \quad \cancel{2x}+3y = -6 \\ & + \ \ \underline{-\cancel{2x}+y = -2}\\ & \qquad \qquad \ 4y = -8\\ & \qquad \qquad \ \ y= -2\end{align*}

Substitute \begin{align*}y\end{align*} into one equation to find \begin{align*}x\end{align*}:

\begin{align*}2x+3(-2) &= -6\\ 2x-6 &= -6\\ 2x &= 0\\ x &= 0\end{align*}

Solution: (0, -2)

### Review

Solve the following systems using linear combinations.

1. .
\begin{align*}4x+2y &= -6\\ -5x-2y &= 4\end{align*}
1. .
\begin{align*}-3x+5y &= -34\\ 3x-y &= 14\end{align*}
1. .
\begin{align*}x+y &= -1\\ x-y &= 21\end{align*}
1. .
\begin{align*}2x+8y &= -4\\ -2x+3y &= 15\end{align*}
1. .
\begin{align*}8x-12y &= 24\\ -3x+12y &= 21\end{align*}
1. .
\begin{align*}x+3y &= -2\\ -x-2y &= 4\end{align*}
1. .
\begin{align*}5x+7y &= 2\\ 5x+3y &= 38\end{align*}
1. .
\begin{align*}12x-2y &= 2\\ 5x-2y &= -5\end{align*}
1. .
\begin{align*}2x+y &= 25\\ x+y &=5\end{align*}
1. .
\begin{align*}\frac{1}{2}x+3y &= -3\\ y &= \frac{1}{2}x-5\end{align*}
1. .
\begin{align*}3x+5y &= 10\\ y &= -3x-10\end{align*}
1. .
\begin{align*}6x+3y &= -3\\ 3y &= -7x+1\end{align*}
1. .
\begin{align*}4x-2y &= 5\\ -4x+2y &= 11\end{align*}
1. .
\begin{align*}9x+2y &= 0\\ -9x-3y &= 0\end{align*}
1. .
\begin{align*}11x+7y &= 12\\ -11x &= 7y -12\end{align*}

Set up and solve a linear system of equations to solve the following word problems.

1. The sum of two numbers is 15 and their difference is 3. Find the two numbers.
2. Jessica and Maria got to the supermarket to buy fruit. Jessica buys 5 apples and 6 oranges and her total before tax is $3.05. Maria buys 7 apples and 6 oranges and her total before tax is$3.55. What is the price of each fruit? Hint: Let \begin{align*}x\end{align*} be the price of one apple and \begin{align*}y\end{align*} be the price of one orange.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 3.6.

### Vocabulary Language: English

Additive inverse

The additive inverse or opposite of a number x is -1(x). A number and its additive inverse always sum to zero.

elimination

The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable.

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