Elimination Method 3

Topics

Algebra 1

Systems of Equations

Elimination Method

Elimination Method 3

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Question 1 of 5

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

$2x+3y=3$

$4x-2y=14$
- $x=$ (3)
  
  $y=$ (-1)
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Elimination Method

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

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

$3)$ solve for the variable that remains

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

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

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

$4x-2y$ $=$ $14$ Equation $2$

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

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

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

$6x+9y$ $=$ $6$ Equation $3$

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

$4x-2y$ $=$ $14$

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

$-8y$ $=$ $8$ $4x-4x$ cancels out

Solve for $y$ from the difference.

$-8y$ $=$ $8$

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

$y$ $=$ $-1$

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

$4x-2$ $y$ $=$ $14$ Equation $2$

$4x-2$ $(-1)$ $=$ $14$ $y=-1$

$4x+2$ $-2$ $=$ $14$ $-2$ Subtract $2$ from both sides

$4x$ $div4$ $=$ $12$ $div4$ Divide both sides by $4$

$x$ $=$ $3$

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

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

$3x+2y=8$

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

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

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

$3)$ solve for the variable that remains

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

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

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

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

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

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

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

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

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

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

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

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

Solve for $y$ from the difference.

$4y$ $=$ $4$

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

$y$ $=$ $1$

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

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

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

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

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

$x$ $=$ $2$

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

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

$3x+5y=6$

$3x-2y=-1$

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

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

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

$3)$ solve for the variable that remains

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

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

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

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

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

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

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

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

Solve for $y$ from the difference.

$7y$ $=$ $7$

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

$y$ $=$ $1$

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

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

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

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

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

$x$ $=$ $1/3$

$x=1/3, y =1$
Question 4 of 5

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

$5x+2y=-3$

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

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

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

$3)$ solve for the variable that remains

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

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

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

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

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

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

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

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

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

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

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

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

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

$\text(Infinite Solutions)$

Question 5 of 5

Solve the following systems of equations by elimination.

$a/4+b=6$

$a/6+2b=8$

$a=$ (12)

$b=$ (3)

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Elimination Method

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

First, label the two equations

$1$ and

$2$ respectively.

$a/4+b$	$=$	$6$	Equation $1$

$a/6+2b$	$=$	$8$	Equation $2$

Next, multiply the values of equation

$1$ by

$4$ and label the product as equation

$3$ .

$a/4+b$	$=$	$6$	Equation $1$

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

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

Also multiply the values of equation

$2$ by

$6$ and label the product as equation

$4$ .

$a/6+2b$	$=$	$8$	Equation $2$

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

Then, subtract equation

$4$ from equation

$3$ .

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

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

Solve for $b$ from the difference.

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

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

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

$a=12, b =3$

Elimination Method 3

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Quiz summary

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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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Great Work!

Elimination Method

Quizzes

$2x+3y$	$=$	$3$	Equation $1$
$(2x+3y)$ $times2$	$=$	$3$ $times2$	Multiply the values of both sides by $3$
$6x+9y$	$=$	$6$	Equation $3$

$4x-2y$	$=$	$14$
$-$ $(4x+6y)$	$=$	$6$

$-8y$	$=$	$8$	$4x-4x$ cancels out

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

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

$3x+2y$	$=$	$8$	Equation $1$
$(3x+2y)$ $times2$	$=$	$8$ $times2$	Multiply the values of both sides by $3$
$6x+4y$	$=$	$16$	Equation $3$

$6x+8y$	$=$	$20$
$-$ $(6x+4y)$	$=$	$16$

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

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

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

$3x+5y$	$=$	$6$
$-$ $(3x-2y)$	$=$	$-1$

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

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

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

$x$	$=$	$1/3$

$5x+2y$	$=$	$-3$	Equation $1$
$(5x+2y)$ $times3$	$=$	$-3$ $times3$	Multiply the values of both sides by $3$
$10x+6y$	$=$	$-6$	Equation $3$