Elimination Method 4

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Algebra 2

Systems of Equations

Elimination Method

Elimination Method 4

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Question 1 of 5

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

$4 x - 3 y = 7$ $4x-3y=7$

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

$4x-3y$ $=$ $7$ Equation $1$

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

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

$4x-3y$ $=$ $7$ Equation $1$

$(4x-3y)$ $times2$ $=$ $7$ $times2$ Multiply the values of both sides by $2$

$8x-6y$ $=$ $14$ Equation $3$

Also multiply the values of equation $2$ by $3$ and label the product as equation $4$ .

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

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

$15x+6y$ $=$ $9$ Equation $4$

Then, add equation $3$ to equation $4$ .

$8x-6y$ $=$ $14$

$+$ $(15x+6y)$ $=$ $9$

$23x$ $=$ $23$ $-6x+6x$ cancels out

Solve for $x$ from the sum.

$23x$ $=$ $23$

$23x$ $div23$ $=$ $23$ $div23$ Divide both sides by $23$

$x$ $=$ $1$

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

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

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

$5+2y$ $-5$ $=$ $3$ $-5$ Subtract $5$ from both sides

$2y$ $div2$ $=$ $-2$ $div2$ Divide both sides by $2$

$y$ $=$ $-1$

$x=1, y =-1$

Question 2 of 5

Solve the following systems of equations by elimination.

$2m+3n=18$

$4m-2n=12$

Write mixed numbers in the format “a b/c”

$x=$ (4 1/2)

$y=$ (3)

Question 3 of 5

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

$6x-y=-2$

$-18x+3y=4$
- 1. $x=2,y=-4$
- 2. $x=-1,y=3$
- 3. $\text(No Solution)$
- 4. $\text(Infinite Solutions)$
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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.

$6x-y$ $=$ $-2$ Equation $1$

$-18x+3y$ $=$ $4$ Equation $2$

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

$6x-y$ $=$ $-2$ Equation $1$

$(6x-y)$ $times3$ $=$ $-2$ $times3$ Multiply the values of both sides by $3$

$18x-3y$ $=$ $-6$ Equation $3$

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

$-18x+3y$ $=$ $4$

$+$ $(18x-3y)$ $=$ $-6$

$0$ $=$ $-2$ $-18x+18x$ and $3y-3y$ cancel out

Since both values with the $x$ and $y$ variables were eliminated, we cannot determine their values with this method.

Therefore, these simultaneous equations have no solution.

$\text(No Solution)$
Question 4 of 5

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

$2x-3y=5$

$4x-6y=14$
- 1. $x=2,y=-3$
- 2. $\text(No Solution)$
- 3. $\text(Infinite Solutions)$
- 4. $x=3,y=2$
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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.

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

$4x-6y$ $=$ $14$ Equation $2$

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

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

$(2x-3y)$ $times2$ $=$ $5$ $times2$ Multiply the values of both sides by $2$

$4x-6y$ $=$ $10$ Equation $3$

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

$4x-6y$ $=$ $14$

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

$0$ $=$ $4$ $4x-4x$ and $-6y-6y$ cancel out

Since both values with the $x$ and $y$ variables were eliminated, we cannot determine their values with this method.

Therefore, these systems of equations have no solution.

$\text(No Solution)$

Question 5 of 5

Solve the following systems of equations by elimination.

$x+y/2=5$

$(x+4y)/3=4$

$x=$ (4)

$y=$ (2)

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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.

$x+y/2$	$=$	$5$	Equation $1$

$(x+4y)/3$	$=$	$4$	Equation $2$

Next, multiply the values of equation

$1$ by

$2$ and label the product as equation

$3$ .

$x+y/2$	$=$	$5$	Equation $1$

$\left(x+\frac{y}{2}\right)\color{#CC0000}{\times2}$	$=$	$5$ $times2$	Multiply the values of both sides by $2$ to cancel the fraction

$2x+y$	$=$	$10$	Equation $3$

Also multiply the values of equation

$2$ by

$3$ and label the product as equation

$4$ .

$(x+4y)/3$	$=$	$4$	Equation $2$

$\left(\frac{x+4y}{3}\right)\color{#CC0000}{\times3}$	$=$	$4$ $times3$	Multiply the values of both sides by $3$ to cancel the fraction
$x+4y$	$=$	$12$	Equation $4$

Now multiply the values of equation

$4$ by

$2$ and label the product as equation

$5$ .

$x+4y$	$=$	$12$	Equation $4$

$(x+4y)$ $times2$	$=$	$12$ $times2$	Multiply the values of both sides by $2$
$2x+8y$	$=$	$24$	Equation $5$

Then, subtract equation

$3$ from equation

$5$ .

$2x+8y$	$=$	$24$
$-$ $(2x+y)$	$=$	$10$

$7y$	$=$	$14$	$2x-2x$ cancels out

Solve for $y$ from the difference.

$7y$	$=$	$14$
$7y$ $div7$	$=$	$14$ $div7$	Divide both sides by $7$
$y$	$=$	$2$

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

$2x+$ $y$	$=$	$10$	Equation $3$
$2x+$ $(2)$	$=$	$10$	$y=2$
$2x+2$ $-2$	$=$	$10$ $-2$	Subtract $2$ from both sides
$2x$ $div2$	$=$	$8$ $div2$	Divide both sides by $2$
$x$	$=$	$4$

$x=4, y =2$

$4m-2n$	$=$	$12$
$-$ $(4m+6n)$	$=$	$36$

$-8n$	$=$	$-24$	$4m-4m$ cancels out

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

$2m+3$ $n$	$=$	$18$	Equation $1$
$2m+3$ $(3)$	$=$	$18$	$n=3$
$2m+9$ $-9$	$=$	$18$ $-9$	Subtract $9$ from both sides
$2m$ $div2$	$=$	$9$ $div2$	Divide both sides by $2$

$m$	$=$	$9/2$

$m$	$=$	$4 1/2$	Simplify

Elimination Method 4

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Elimination Method

2. Question

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Elimination Method

3. Question

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Elimination Method

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Elimination Method

5. Question

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Elimination Method

Quizzes

$4x-3y$	$=$	$7$	Equation $1$
$(4x-3y)$ $times2$	$=$	$7$ $times2$	Multiply the values of both sides by $2$
$8x-6y$	$=$	$14$	Equation $3$

$5x+2y$	$=$	$3$	Equation $2$
$(5x+2y)$ $times3$	$=$	$3$ $times3$	Multiply the values of both sides by $3$
$15x+6y$	$=$	$9$	Equation $4$

$8x-6y$	$=$	$14$
$+$ $(15x+6y)$	$=$	$9$

$23x$	$=$	$23$	$-6x+6x$ cancels out

$23x$	$=$	$23$
$23x$ $div23$	$=$	$23$ $div23$	Divide both sides by $23$
$x$	$=$	$1$

$5$ $x$ $+2y$	$=$	$3$	Equation $2$
$5$ $(1)$ $+2y$	$=$	$3$	$x=1$
$5+2y$ $-5$	$=$	$3$ $-5$	Subtract $5$ from both sides
$2y$ $div2$	$=$	$-2$ $div2$	Divide both sides by $2$
$y$	$=$	$-1$

$2m+3n$	$=$	$18$	Equation $1$
$(2m+3n)$ $times2$	$=$	$18$ $times2$	Multiply the values of both sides by $3$
$4m+6n$	$=$	$36$	Equation $3$

$6x-y$	$=$	$-2$	Equation $1$
$(6x-y)$ $times3$	$=$	$-2$ $times3$	Multiply the values of both sides by $3$
$18x-3y$	$=$	$-6$	Equation $3$

$-18x+3y$	$=$	$4$
$+$ $(18x-3y)$	$=$	$-6$

$0$	$=$	$-2$	$-18x+18x$ and $3y-3y$ cancel out

$2x-3y$	$=$	$5$	Equation $1$
$(2x-3y)$ $times2$	$=$	$5$ $times2$	Multiply the values of both sides by $2$
$4x-6y$	$=$	$10$	Equation $3$

$4x-6y$	$=$	$14$
$-$ $(4x-6y)$	$=$	$10$

$0$	$=$	$4$	$4x-4x$ and $-6y-6y$ cancel out