Elimination Method 2

Topics

Precalculus

Systems of Equations

Elimination Method

Elimination Method 2

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Question 1 of 5

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

$3x+y=5$

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

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

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

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

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

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

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

Solve for $x$ from the difference.

$2x$ $=$ $2$

$2x$ $div2$ $=$ $2$ $div2$ Divide both sides by $2$

$x$ $=$ $1$

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

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

$1$ $+y$ $=$ $3$ $x=1$

$1+y$ $-1$ $=$ $3$ $-1$ Subtract $1$ from both sides

$y$ $=$ $2$

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

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

$5x+y=-6$

$x+2y=24$
- $x=$ (-4)
  
  $y=$ (14)
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In the elimination method you either add or subtract the equations to get the value of $x$ and $y$

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

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

$x+2y$ $=$ $24$ Equation $2$

Multiply Equation $1$ by $2$ .

$5x+y$ $=$ $-6$

$(5x+y)$ $xx 2$ $=$ $-6$ $xx 2$

$10x+2y$ $=$ $-12$ Simplify

Next, Subtract equation $2$ from the transformed equation.

$10x+2y$ $=$ $-12$

$x+2y$ $=$ $24$

$9x$ $=$ $-36$ $2y-2y$ cancels out

Solve for $x$ .

$9x$ $=$ $-36$

$x$ $=$ $-4$ Divide both sides by $9$

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

$x$ $+2y$ $=$ $24$ Equation $2$

$-4$ $+2y$ $=$ $24$ $x=-4$

$-4+2y$ $+4$ $=$ $24$ $+4$ Add $4$ to both sides

$2y$ $=$ $28$

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

$x=-4, y =14$
Question 3 of 5

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

$3a-4b=24$

$4a-2b=12$
- $a=$ (0)
  
  $b=$ (-6)
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In the elimination method you either add or subtract the equations to get the value of $a$ and $b$

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

$3a-4b$ $=$ $24$ Equation $1$

$4a-2b$ $=$ $12$ Equation $2$

Multiply Equation $2$ by $2$ .

$4a-2b$ $=$ $12$

$(4a-2b)$ $xx 2$ $=$ $12$ $xx 2$

$8a-4b$ $=$ $24$ Simplify

Subtract equation $1$ from the transformed equation.

$8a-4b$ $=$ $24$

$3a-4b$ $=$ $24$

$5a$ $=$ $0$ $-4b-(-4b)$ cancels out

Solve for $a$ .

$5a$ $=$ $0$

$a$ $=$ $0$ Divide both sides by $5$

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

$3$ $a$ $-4b$ $=$ $24$ Equation $1$

$3$ $(0)$ $-4b$ $=$ $24$ $a=0$

$-4b$ $=$ $24$

$b$ $=$ $-6$ Divide both sides by $-4$

$a=0, b=-6$
Question 4 of 5

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

$a-5b=8$

$2a-3b=9$
- $x=$ (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.

$a-5b$ $=$ $8$ Equation $1$

$2a-3b$ $=$ $9$ Equation $2$

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

$a-5b$ $=$ $8$ Equation $1$

$(a-5b)$ $times2$ $=$ $8$ $times2$ Multiply the values of both sides by $2$

$2a-10b$ $=$ $16$ Equation $3$

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

$2a-10b$ $=$ $16$

$-$ $(2a-3b)$ $=$ $9$

$-7b$ $=$ $7$ $2a-2a$ cancels out

Solve for $b$ from the difference.

$-7b$ $=$ $7$

$-7b$ $div(-7)$ $=$ $7$ $div(-7)$ Divide both sides by $-7$

$b$ $=$ $-1$

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

$a-5$ $b$ $=$ $8$ Equation $1$

$a-5$ $(-1)$ $=$ $8$ $b=-1$

$a+5$ $-5$ $=$ $8$ $-5$ Subtract $5$ from both sides

$a$ $=$ $3$

$a=3, y =-1$

Question 5 of 5

Solve the following systems of equations by elimination.

$3x-5y=11$

$2x-y=5$

$x=$ (2)

$y=$ (-1)

$10x-5y$	$=$	$25$
$-$ $(3x-5y)$	$=$	$11$

$7x$	$=$	$14$	$-5y-(-5y)$ cancels out

$3$ $x$ $-5y$	$=$	$11$	Equation $1$
$3$ $(2)$ $-5y$	$=$	$11$	$x=2$
$6-5y$ $-6$	$=$	$11$ $-6$	Subtract $6$ from both sides
$-5y$ $div(-5)$	$=$	$5$ $div(-5)$	Divide both sides by $-5$
$y$	$=$	$-1$

Elimination Method 2

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Elimination Method

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Elimination Method

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Elimination Method

Quizzes

$3x+y$	$=$	$5$
$-$ $(x+y)$	$=$	$3$

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

$2x$	$=$	$2$
$2x$ $div2$	$=$	$2$ $div2$	Divide both sides by $2$
$x$	$=$	$1$

$x$ $+y$	$=$	$3$	Equation $2$
$1$ $+y$	$=$	$3$	$x=1$
$1+y$ $-1$	$=$	$3$ $-1$	Subtract $1$ from both sides
$y$	$=$	$2$

$5x+y$	$=$	$-6$
$(5x+y)$ $xx 2$	$=$	$-6$ $xx 2$
$10x+2y$	$=$	$-12$	Simplify

$10x+2y$	$=$	$-12$
$x+2y$	$=$	$24$

$9x$	$=$	$-36$	$2y-2y$ cancels out

$x$ $+2y$	$=$	$24$	Equation $2$
$-4$ $+2y$	$=$	$24$	$x=-4$
$-4+2y$ $+4$	$=$	$24$ $+4$	Add $4$ to both sides
$2y$	$=$	$28$
$y$	$=$	$14$	Divide both sides by $2$

$4a-2b$	$=$	$12$
$(4a-2b)$ $xx 2$	$=$	$12$ $xx 2$
$8a-4b$	$=$	$24$	Simplify

$8a-4b$	$=$	$24$
$3a-4b$	$=$	$24$

$5a$	$=$	$0$	$-4b-(-4b)$ cancels out

$3$ $a$ $-4b$	$=$	$24$	Equation $1$
$3$ $(0)$ $-4b$	$=$	$24$	$a=0$
$-4b$	$=$	$24$
$b$	$=$	$-6$	Divide both sides by $-4$

$a-5b$	$=$	$8$	Equation $1$
$(a-5b)$ $times2$	$=$	$8$ $times2$	Multiply the values of both sides by $2$
$2a-10b$	$=$	$16$	Equation $3$

$2a-10b$	$=$	$16$
$-$ $(2a-3b)$	$=$	$9$

$-7b$	$=$	$7$	$2a-2a$ cancels out

$-7b$	$=$	$7$
$-7b$ $div(-7)$	$=$	$7$ $div(-7)$	Divide both sides by $-7$
$b$	$=$	$-1$

$a-5$ $b$	$=$	$8$	Equation $1$
$a-5$ $(-1)$	$=$	$8$	$b=-1$
$a+5$ $-5$	$=$	$8$ $-5$	Subtract $5$ from both sides
$a$	$=$	$3$

$2x-y$	$=$	$5$	Equation $2$
$(2x-y)$ $times5$	$=$	$5$ $times5$	Multiply the values of both sides by $5$
$10x-5y$	$=$	$25$	Equation $3$

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