Elimination Method 3

Question 2 of 5

2. Question
Solve the following systems of equations by elimination.

$3 x + 2 y = 8$ $3x+2y=8$

$6 x + 8 y = 20$ $6x+8y=20$
- $x =$ $x=$ (2)
  
  $y =$ $y=$ (1)
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Elimination Method

$1)$ $1)$ make sure a variable has same coefficients on the 2 equations

$2)$ $2)$ add or subtract the equations so that one variable is cancelled

$3)$ $3)$ solve for the variable that remains

$4)$ $4)$ substitute known value to one of the equations to solve for the other variable

First, label the two equations $1$ $1$ and $2$ $2$ respectively.

$3 x + 2 y$ $3x+2y$ $=$ $=$ $8$ $8$ Equation $1$ $1$

$6 x + 8 y$ $6x+8y$ $=$ $=$ $20$ $20$ Equation $2$ $2$

Next, multiply the values of equation $1$ $1$ by $2$ $2$ and label the product as equation $3$ $3$ .

$3 x + 2 y$ $3x+2y$ $=$ $=$ $8$ $8$ Equation $1$ $1$

$(3 x + 2 y)$ $(3x+2y)$ $\times 2$ $times2$ $=$ $=$ $8$ $8$ $\times 2$ $times2$ Multiply the values of both sides by $3$ $3$

$6 x + 4 y$ $6x+4y$ $=$ $=$ $16$ $16$ Equation $3$ $3$

Then, subtract equation $3$ $3$ from equation $2$ $2$ .

$6 x + 8 y$ $6x+8y$ $=$ $=$ $20$ $20$

$-$ $-$ $(6 x + 4 y)$ $(6x+4y)$ $=$ $=$ $16$ $16$

$4 y$ $4y$ $=$ $=$ $4$ $4$ $6 x - 6 x$ $6x-6x$ cancels out

Solve for $y$ $y$ from the difference.

$4 y$ $4y$ $=$ $=$ $4$ $4$

$4 y$ $4y$ $\div 4$ $div4$ $=$ $=$ $4$ $4$ $\div 4$ $div4$ Divide both sides by $4$ $4$

$y$ $y$ $=$ $=$ $1$ $1$

Now, substitute the value of $y$ $y$ into any of the two equations.

$6 x + 8$ $6x+8$ $y$ $y$ $=$ $=$ $20$ $20$ Equation $2$ $2$

$6 x + 8$ $6x+8$ $(1)$ $(1)$ $=$ $=$ $20$ $20$ $y = 1$ $y=1$

$6 x + 8$ $6x+8$ $- 8$ $-8$ $=$ $=$ $20$ $20$ $- 8$ $-8$ Subtract $8$ $8$ from both sides

$6 x$ $6x$ $\div 6$ $div6$ $=$ $=$ $12$ $12$ $\div 6$ $div6$ Divide both sides by $6$ $6$

$x$ $x$ $=$ $=$ $2$ $2$

$x = 2, y = 1$ $x=2, y =1$
Question 3 of 5

3. Question
Solve the following systems of equations by elimination.

$3 x + 5 y = 6$ $3x+5y=6$

$3 x - 2 y = - 1$ $3x-2y=-1$

Write fractions in the format “a/b”
- $x =$ $x=$ (1/3)
  
  $y =$ $y=$ (1)
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Elimination Method

$1)$ $1)$ make sure a variable has same coefficients on the 2 equations

$2)$ $2)$ add or subtract the equations so that one variable is cancelled

$3)$ $3)$ solve for the variable that remains

$4)$ $4)$ substitute known value to one of the equations to solve for the other variable

First, label the two equations $1$ $1$ and $2$ $2$ respectively.

$3 x + 5 y$ $3x+5y$ $=$ $=$ $6$ $6$ Equation $1$ $1$

$3 x - 2 y$ $3x-2y$ $=$ $=$ $- 1$ $-1$ Equation $2$ $2$

Next, subtract equation $2$ $2$ from equation $1$ $1$ .

$3 x + 5 y$ $3x+5y$ $=$ $=$ $6$ $6$

$-$ $-$ $(3 x - 2 y)$ $(3x-2y)$ $=$ $=$ $- 1$ $-1$

$7 y$ $7y$ $=$ $=$ $7$ $7$ $3 x - 3 x$ $3x-3x$ cancels out

Solve for $y$ $y$ from the difference.

$7 y$ $7y$ $=$ $=$ $7$ $7$

$7 y$ $7y$ $\div 7$ $div7$ $=$ $=$ $7$ $7$ $\div 7$ $div7$ Divide both sides by $7$ $7$

$y$ $y$ $=$ $=$ $1$ $1$

Now, substitute the value of $y$ $y$ into any of the two equations.

$3 x + 5$ $3x+5$ $y$ $y$ $=$ $=$ $6$ $6$ Equation $1$ $1$

$3 x + 5$ $3x+5$ $(1)$ $(1)$ $=$ $=$ $6$ $6$ $y = 1$ $y=1$

$3 x + 5$ $3x+5$ $- 5$ $-5$ $=$ $=$ $6$ $6$ $- 5$ $-5$ Subtract $5$ $5$ from both sides

$3 x$ $3x$ $\div 3$ $div3$ $=$ $=$ $1$ $1$ $\div 3$ $div3$ Divide both sides by $3$ $3$

$x$ $x$ $=$ $=$ $\frac{1}{3}$ $1/3$

$x = \frac{1}{3}, y = 1$ $x=1/3, y =1$
Question 4 of 5

4. Question
Solve the following systems of equations by elimination.

$5 x + 2 y = - 3$ $5x+2y=-3$

$- 10 x - 4 y = 6$ $-10x-4y=6$
- 1. $No Solution$ $\text(No Solution)$
- 2. $x = - 2, y = - 1$ $x=-2,y=-1$
- 3. $x = - 1, y = 2$ $x=-1,y=2$
- 4. $Infinite Solutions$ $\text(Infinite Solutions)$
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Elimination Method

$1)$ $1)$ make sure a variable has same coefficients on the 2 equations

$2)$ $2)$ add or subtract the equations so that one variable is cancelled

$3)$ $3)$ solve for the variable that remains

$4)$ $4)$ substitute known value to one of the equations to solve for the other variable

First, label the two equations $1$ $1$ and $2$ $2$ respectively.

$5 x + 2 y$ $5x+2y$ $=$ $=$ $- 3$ $-3$ Equation $1$ $1$

$- 10 x - 4 y$ $-10x-4y$ $=$ $=$ $6$ $6$ Equation $2$ $2$

Next, multiply the values of equation $1$ $1$ by $2$ $2$ and label the product as equation $3$ $3$ .

$5 x + 2 y$ $5x+2y$ $=$ $=$ $- 3$ $-3$ Equation $1$ $1$

$(5 x + 2 y)$ $(5x+2y)$ $\times 3$ $times3$ $=$ $=$ $- 3$ $-3$ $\times 3$ $times3$ Multiply the values of both sides by $3$ $3$

$10 x + 6 y$ $10x+6y$ $=$ $=$ $- 6$ $-6$ Equation $3$ $3$

Next, subtract equation $3$ $3$ from equation $2$ $2$ .

$- 10 x - 4 y$ $-10x-4y$ $=$ $=$ $6$ $6$

$-$ $-$ $(10 x + 4 y)$ $(10x+4y)$ $=$ $=$ $- 6$ $-6$

Applying the rule of subtracting integers where we change the signs of each value on the subtrahend, we will be getting $- 10 x - 4 y = 6$ $-10x-4y=6$ as the subtrahend, which is the same as equation $2$ $2$ .

If the systems of equations have the same linear equations, there will be infinite solutions.

$Infinite Solutions$ $\text(Infinite Solutions)$

Question 5 of 5

Solve the following systems of equations by elimination.

\frac{a}{4} + b = 6

$a/4+b=6$

\frac{a}{6} + 2 b = 8

$a/6+2b=8$

$a =$ $a=$ (12)

$b =$ $b=$ (3)

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Elimination Method

$1)$ $1)$ make sure a variable has same coefficients on the 2 equations
$2)$ $2)$ add or subtract the equations so that one variable is cancelled
$3)$ $3)$ solve for the variable that remains
$4)$ $4)$ substitute known value to one of the equations to solve for the other variable

First, label the two equations

1

$1$ and

2

$2$ respectively.

$\frac{a}{4} + b$ $a/4+b$	$=$ $=$	$6$ $6$	Equation $1$ $1$

$\frac{a}{6} + 2 b$ $a/6+2b$	$=$ $=$	$8$ $8$	Equation $2$ $2$

Next, multiply the values of equation

1

$1$ by

4

$4$ and label the product as equation

3

$3$ .

$\frac{a}{4} + b$ $a/4+b$	$=$ $=$	$6$ $6$	Equation $1$ $1$

$\left(\frac{a}{4}+b\right)\color{#CC0000}{\times4}$	$=$ $=$	$6$ $6$ $\times 4$ $times4$	Multiply the values of both sides by $4$ $4$ to cancel the fraction

$a + 4 b$ $a+4b$	$=$ $=$	$24$ $24$	Equation $3$ $3$

Also multiply the values of equation

2

$2$ by

6

$6$ and label the product as equation

4

$4$ .

$\frac{a}{6} + 2 b$ $a/6+2b$	$=$ $=$	$8$ $8$	Equation $2$ $2$

$\left(\frac{a}{6}+2b\right)\color{#CC0000}{\times6}$	$=$ $=$	$8$ $8$ $\times 6$ $times6$	Multiply the values of both sides by $6$ $6$ to cancel the fraction
$a + 12 b$ $a+12b$	$=$ $=$	$48$ $48$	Equation $4$ $4$

Then, subtract equation

4

$4$ from equation

3

$3$ .

$a + 4 b$ $a+4b$	$=$ $=$	$24$ $24$
$-$ $-$ $(a + 12 b)$ $(a+12b)$	$=$ $=$	$48$ $48$

$- 8 b$ $-8b$	$=$ $=$	$- 24$ $-24$	$a - a$ $a-a$ cancels out

Solve for $b$ $b$ from the difference.

$- 8 b$ $-8b$	$=$ $=$	$- 24$ $-24$
$- 8 b$ $-8b$ $\div (- 8)$ $div(-8)$	$=$ $=$	$- 24$ $-24$ $\div (- 8)$ $div(-8)$	Divide both sides by $- 8$ $-8$
$b$ $b$	$=$ $=$	$3$ $3$

Now, substitute the value of $b$ $b$ into any of the four equations.

$a + 4$ $a+4$ $b$ $b$	$=$ $=$	$24$ $24$	Equation $3$ $3$
$a + 4$ $a+4$ $(3)$ $(3)$	$=$ $=$	$24$ $24$	$b = 3$ $b=3$
$a + 12$ $a+12$ $- 12$ $-12$	$=$ $=$	$24$ $24$ $- 12$ $-12$	Subtract $12$ $12$ from both sides
$a$ $a$	$=$ $=$	$12$ $12$

a = 12, b = 3

$a=12, b =3$

$4 x - 2$ $4x-2$ $y$ $y$	$=$ $=$	$14$ $14$	Equation $2$ $2$
$4 x - 2$ $4x-2$ $(- 1)$ $(-1)$	$=$ $=$	$14$ $14$	$y = - 1$ $y=-1$
$4 x + 2$ $4x+2$ $- 2$ $-2$	$=$ $=$	$14$ $14$ $- 2$ $-2$	Subtract $2$ $2$ from both sides
$4 x$ $4x$ $\div 4$ $div4$	$=$ $=$	$12$ $12$ $\div 4$ $div4$	Divide both sides by $4$ $4$
$x$ $x$	$=$ $=$	$3$ $3$

Elimination Method 3

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1. Question

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Elimination Method

2. Question

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Elimination Method

3. Question

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Elimination Method

4. Question

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Elimination Method

5. Question

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Elimination Method

Quizzes

$2 x + 3 y$ $2x+3y$	$=$ $=$	$3$ $3$	Equation $1$ $1$
$4 x - 2 y$ $4x-2y$	$=$ $=$	$14$ $14$	Equation $2$ $2$

$- 8 y$ $-8y$	$=$ $=$	$8$ $8$
$- 8 y$ $-8y$ $\div (- 8)$ $div(-8)$	$=$ $=$	$8$ $8$ $\div (- 8)$ $div(-8)$	Divide both sides by $- 8$ $-8$
$y$ $y$	$=$ $=$	$- 1$ $-1$

$3 x + 2 y$ $3x+2y$	$=$ $=$	$8$ $8$	Equation $1$ $1$
$6 x + 8 y$ $6x+8y$	$=$ $=$	$20$ $20$	Equation $2$ $2$

$4 y$ $4y$	$=$ $=$	$4$ $4$
$4 y$ $4y$ $\div 4$ $div4$	$=$ $=$	$4$ $4$ $\div 4$ $div4$	Divide both sides by $4$ $4$
$y$ $y$	$=$ $=$	$1$ $1$

$6 x + 8$ $6x+8$ $y$ $y$	$=$ $=$	$20$ $20$	Equation $2$ $2$
$6 x + 8$ $6x+8$ $(1)$ $(1)$	$=$ $=$	$20$ $20$	$y = 1$ $y=1$
$6 x + 8$ $6x+8$ $- 8$ $-8$	$=$ $=$	$20$ $20$ $- 8$ $-8$	Subtract $8$ $8$ from both sides
$6 x$ $6x$ $\div 6$ $div6$	$=$ $=$	$12$ $12$ $\div 6$ $div6$	Divide both sides by $6$ $6$
$x$ $x$	$=$ $=$	$2$ $2$

$3 x + 5 y$ $3x+5y$	$=$ $=$	$6$ $6$	Equation $1$ $1$
$3 x - 2 y$ $3x-2y$	$=$ $=$	$- 1$ $-1$	Equation $2$ $2$

$7 y$ $7y$	$=$ $=$	$7$ $7$
$7 y$ $7y$ $\div 7$ $div7$	$=$ $=$	$7$ $7$ $\div 7$ $div7$	Divide both sides by $7$ $7$
$y$ $y$	$=$ $=$	$1$ $1$

$3 x + 5$ $3x+5$ $y$ $y$	$=$ $=$	$6$ $6$	Equation $1$ $1$
$3 x + 5$ $3x+5$ $(1)$ $(1)$	$=$ $=$	$6$ $6$	$y = 1$ $y=1$
$3 x + 5$ $3x+5$ $- 5$ $-5$	$=$ $=$	$6$ $6$ $- 5$ $-5$	Subtract $5$ $5$ from both sides
$3 x$ $3x$ $\div 3$ $div3$	$=$ $=$	$1$ $1$ $\div 3$ $div3$	Divide both sides by $3$ $3$

$x$ $x$	$=$ $=$	$\frac{1}{3}$ $1/3$

$5 x + 2 y$ $5x+2y$	$=$ $=$	$- 3$ $-3$	Equation $1$ $1$
$- 10 x - 4 y$ $-10x-4y$	$=$ $=$	$6$ $6$	Equation $2$ $2$