Elimination Method 4

Topics

Precalculus

Systems of Equations

Elimination Method

Elimination Method 4

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Question 1 of 5

Solve the following systems of equations by elimination.

4 x - 3 y = 7

$4x-3y=7$

5 x + 2 y = 3

$5x+2y=3$

$x =$ $x=$ (1)

$y =$ $y=$ (-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.

$4 x - 3 y$ $4x-3y$	$=$ $=$	$7$ $7$	Equation $1$ $1$
$5 x + 2 y$ $5x+2y$	$=$ $=$	$3$ $3$	Equation $2$ $2$

Next, multiply the values of equation

1

$1$ by

2

$2$ and label the product as equation

3

$3$ .

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

Also multiply the values of equation

2

$2$ by

3

$3$ and label the product as equation

4

$4$ .

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

Then, add equation

3

$3$ to equation

4

$4$ .

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

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

Solve for $x$ $x$ from the sum.

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

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

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

x = 1, y = - 1

$x=1, y =-1$

Question 2 of 5

Solve the following systems of equations by elimination.

2 m + 3 n = 18

$2m+3n=18$

4 m - 2 n = 12

$4m-2n=12$

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

$x =$ $x=$ (4 1/2)

$y =$ $y=$ (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.

$2 m + 3 n$ $2m+3n$	$=$ $=$	$18$ $18$	Equation $1$ $1$
$4 m - 2 n$ $4m-2n$	$=$ $=$	$12$ $12$	Equation $2$ $2$

Next, multiply the values of equation

1

$1$ by

2

$2$ and label the product as equation

3

$3$ .

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

Then, subtract equation

3

$3$ from equation

2

$2$ .

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

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

Solve for $n$ $n$ from the difference.

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

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

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

$m$ $m$	$=$ $=$	$\frac{9}{2}$ $9/2$

$m$ $m$	$=$ $=$	$4 \frac{1}{2}$ $4 1/2$	Simplify

m = 4 \frac{1}{2}, n = 3

$m=4 1/2, n =3$

Question 3 of 5

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

$6 x - y = - 2$ $6x-y=-2$

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

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

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

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

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

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

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

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

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

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

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

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

Therefore, these simultaneous equations have no solution.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Therefore, these systems of equations have no solution.

$No Solution$ $\text(No Solution)$

Question 5 of 5

Solve the following systems of equations by elimination.

x + \frac{y}{2} = 5

$x+y/2=5$

\frac{x + 4 y}{3} = 4

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

$x =$ $x=$ (4)

$y =$ $y=$ (2)

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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.

$x + \frac{y}{2}$ $x+y/2$	$=$ $=$	$5$ $5$	Equation $1$ $1$

$\frac{x + 4 y}{3}$ $(x+4y)/3$	$=$ $=$	$4$ $4$	Equation $2$ $2$

Next, multiply the values of equation

1

$1$ by

2

$2$ and label the product as equation

3

$3$ .

$x + \frac{y}{2}$ $x+y/2$	$=$ $=$	$5$ $5$	Equation $1$ $1$

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

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

Also multiply the values of equation

2

$2$ by

3

$3$ and label the product as equation

4

$4$ .

$\frac{x + 4 y}{3}$ $(x+4y)/3$	$=$ $=$	$4$ $4$	Equation $2$ $2$

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

Now multiply the values of equation

4

$4$ by

2

$2$ and label the product as equation

5

$5$ .

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

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

Then, subtract equation

3

$3$ from equation

5

$5$ .

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

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

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

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

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

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

x = 4, y = 2

$x=4, y =2$

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

4. Question

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Elimination Method

5. Question

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Elimination Method

Quizzes

$6 x - y$ $6x-y$	$=$ $=$	$- 2$ $-2$	Equation $1$ $1$
$- 18 x + 3 y$ $-18x+3y$	$=$ $=$	$4$ $4$	Equation $2$ $2$

$2 x - 3 y$ $2x-3y$	$=$ $=$	$5$ $5$	Equation $1$ $1$
$4 x - 6 y$ $4x-6y$	$=$ $=$	$14$ $14$	Equation $2$ $2$