Elimination Method 3

Question 2 of 5

2. Question
Solve the following simultaneous 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 $2$ $2$ .

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

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

$4y$ $=$ $4$ $6x-6x$ 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 simultaneous 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 simultaneous equations by elimination.

$5x+2y=-3$

$-10x-4y=6$
- 1. $x=-1,y=2$
- 2. $x=-2,y=-1$
- 3. $\text(Infinite Solutions)$
- 4. $\text(No Solution)$
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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 simultaneous 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
$a$	$=$	$12$

$a=12, b =3$

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

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

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

$4 x - 2$ $4x-2$ $y$ $y$	$=$ $=$	$14$ $14$	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 $4$ $4$
$x$ $x$	$=$ $=$	$3$ $3$

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Quiz summary

Information

1. Question

Hint

Correct!

Elimination Method

2. Question

Hint

Fantastic!

Elimination Method

3. Question

Hint

Nice Job!

Elimination Method

4. Question

Hint

Keep Going!

Elimination Method

5. Question

Hint

Great Work!

Elimination Method

Quizzes

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

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

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

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