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Solving Systems by Multiplying One Equation

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Solving Systems by Multiplying One Equation to Cancel a Variable

Mattie wants to plant some flowers in her yard. She has space for 15 plants. She buys pansies and daisies at her local garden center. The pansies are each $2.75 and the daisies are each $2.00. How many of each does she buy if she spends a total of $35.25?

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Khan Academy: Solving systems by elimination 2

Guidance

It is not always the case that the coefficients of one variable in a system of linear equations are the same or opposites. When this is not the case, we may be able to multiply one of the equations by a constant so that we have opposite coefficients of one variable and can proceed to solve the system as we have previously done.

Example A

Solve the system using linear combinations:

4x+y &= 0\\x- 3y &= 26

Solution: Here we have coefficients of y that are opposite signs (one is positive and one is negative). We can get opposite values if we multiply the first equation by 3. Be careful; make sure you multiply the entire equation, including the constant, by 3:

3(4x+y &= 0)\\12x+3y &= 0

Now we can use this new equation in our system to eliminate y and solve for x :

& \quad \ 12x+\cancel{3y}=0\\& + \ \underline{\;\;\; x-\cancel{3y}=26}\\& \qquad \quad \ 13x = 26\\& \qquad \quad \quad \ x=2

Now, find y :

4(2)+y &= 0\\8+y &= 0\\y &= -8

Solution: (2, -8)

Check your answer:

4(2)+(-8) &= 8-8=0\\(2)-3(-8) &= 2+24=26

^* Note that we could have used the other equation to find y in the final step.

^{**} From the beginning, we could have multiplied the second equation by -4 instead to cancel out the x variable.

Example B

Solve the system using linear combinations:

2x+5y &=1\\y &= -3x+21

Solution: For this system, we must first rewrite the second equation in standard form so that we can see how the coefficients compare. If we add 3x to both sides we get:

2x+5y &=1\\3x+y &=21

Now, we can see that if we multiply the second equation by -5, the coefficients of y will be opposites.

& \qquad \qquad 2x+\cancel{5y}=1\\& \quad + \ \ \underline{-15x-\cancel{5y}=-105}\\& \qquad \qquad \ -13x = -104\\& \qquad \qquad \qquad \ x = 8

Now, find y :

y &= -3(8)+21\\y &= -24+21\\ y &= -3

Solution: (8, -3)

Check your answer:

2(8)+5(-3) &= 16-15 = 1\\-3=-3(8)+21 &= -24+21=-3

Example C

Solve the system using linear combinations:

4x-6y &= -12\\y &= \frac{2}{3}x+2

Solution: Again, we need to rearrange the second equation in this system to get it in standard form. We can do this by subtracting \frac{2}{3}x on both sides to get the following system:

4x-6y &= -12\\- \frac{2}{3}x+y &= 2

Multiply the second equation by 6 to eliminate y :

& \quad 6\left(- \frac{2}{3}x+y =2\right)\\& \quad \ \ -4x+6y = 12

And add it to the first equation.

& \qquad \quad 4x-\cancel{6y}= -12\\& \ \ + \ \underline{-4x+\cancel{6y}=12\;\;\;}\\& \qquad \qquad \quad 0x=0\\& \qquad \qquad \quad \ 0=0

Here, both variables were eliminated and we wound up with 0 = 0. Recall that this is a true statement and thus this system has infinite solutions.

Intro Problem Revisit The system of linear equations represented by this situation is:

p + d &= 15\\2.75p + 2d &= 35.25

If we multiply the first of these two equations by -2 , we get a new system of linear equations:

-2p - 2d &= -30\\2.75p + 2d &= 35.25

Now we can add these two equations to cancel out the d variable. When we do so, we get:

0.75p = 5.25 or p = 7

Finally, we can substitute p = 7 into either of our original equations to get the value of d .

7 + d = 15 or d = 8

Therefore Mattie buys 7 pansies and 8 daisies.

Guided Practice

Solve the following systems using linear combinations.

1. 3x+12y &= -3\\-x-5y &= 0

2. 0.75x+5y &= 0\\0.25x-9y &= 0

3. x-3y &= 5\\y &= \frac{1}{3}x+8

Answers

1. In this problem we can just multiply the second equation by 3 to get coefficients of x which are opposites: 3(-x-5y=0)\Rightarrow -3x-15y =0

&\ \ \cancel{3x}+12y=-3\\&\underline{-\cancel{3x}-15y=0\;\;\;}\\& \qquad \ -3y=-3\\& \qquad \qquad y=1

Now we can find x :

3x+12(1) &= -3\\3x+12 &= -3\\3x &= -15\\x &= -5

Solution: (-5, 1)

2. For this system, we need to multiply the second equation by -3 to get coefficients of x which are opposites: -3(0.25x-9y=0) \Rightarrow -0.75x+27y=0

& \quad \cancel{0.75x}+5y=0\\& - \underline{\cancel{0.75x}-9y=0}\\& \qquad \quad \ -4y=0\\& \qquad \qquad \quad y=0

Now we can find x :

0.75x+5(0) &= 0\\0.75x &= 0\\x &= 0

Solution: (0, 0)

3. This time we need to rewrite the second equation in standard form:

x-3y &= 5\\- \frac{1}{3}x+y &= 8.

Now we can multiply the second equation by 3 to get coefficients of x that are opposites:

3 \left(-\frac{1}{3}x+y=8\right) \Rightarrow -x+3y=24 , Now our system is:

x-3y &= 5\\-x+3y &= 24

When we add these equations together, both variables are eliminated and the result is 0 = 29, which is an untrue statement. Therefore, this system has no solution.

Practice

Solve the following systems using linear combinations.

  1. .
x-7y &= 27\\2x+y &= 9
  1. .
x+3y &= 31\\3x-5y &= -5
  1. .
10x+y &= -6\\-7x-5y &= -13
  1. .
2x+4y &= 18\\x-5y &= 9
  1. .
2x+6y &= 8\\3x+2y &= -23
  1. .
12x-y &= 2\\2x+5y &= 21
  1. .
2x+4y &= 24\\-3x-2y &= -26
  1. .
3x+2y &= 19\\5x+4y &= 23
  1. .
3x-9y &= 13\\x-3y &= 7
  1. .
8x+2y &= -4\\3y &= -16x+2
  1. .
3x+2y &= -3\\-6x-5y &= 4
  1. .
10x+6y &= -24\\y &= - \frac{5}{3}x-4
  1. .
\frac{1}{3}x-\frac{2}{3}y &= -8\\\frac{1}{2}x-\frac{1}{3}y &= 12
  1. .
6x-10y &= -8\\y &= - \frac{3}{5}x
  1. .
4x-14y &= -52\\y &= \frac{2}{7}x+3

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

  1. Lia is making a mixture of Chlorine and water in her science class. She needs to make 13 ml of a 60% chlorine solution from a solution that is 35% chlorine and a second solution which is 75% chlorine. How many milliliters of each solution does she need?
  2. Chelsea and Roberto each sell baked goods for their club’s fundraiser. Chelsea sells 13 cookies and 7 brownies and collects a total of $11.75. Roberto sells 10 cookies and 14 brownies and collects a total of $15.50. How much did they charge for the cookies and the brownies?

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