3 Variable Systems of Equations – Elimination Method

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Question 1 of 5

1. Question

Solve the following systems of equations by elimination.

`3x-y+2z=5`

`4x+2y-z=6`

`5x-3y+z=1`

`x=` (1)

`y=` (2)

`z=` (2)

Hint

Help Video

Correct

Nice Job!

Incorrect

Elimination Method

`1)` make sure 2 equations have a variable with same coefficients but opposite signs
`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 equations `1`, `2` and `3` respectively.

`3x-y+2z`	`=`	`5`	Equation `1`
`4x+2y-z`	`=`	`6`	Equation `2`
`5x-3y+z`	`=`	`1`	Equation `3`

Next, add equation `2` to equation `3` since their variable `z` both have `1` as coefficient but with opposite signs. Label the result as equation `A`.

`4x+2y-z`	`=`	`6`
`+` `5x-3y+z`	`=`	`1`	`-z+z` cancels out

`9x-y`	`=`	`7`	Equation `A`

Next, multiply the values of equation `2` by `2` and add the product to equation `1`. Label the result as equation `B`.

`2xx``(4x+2y-z)`	`=`	`2xx``6`	Multiply equation `2` by `2`
`8x+4y-2z`	`=`	`12`

`8x+4y-2z`	`=`	`12`
`+` `3x-y+2z`	`=`	`5`	`+2z-2z` cancels out

`11x+3y`	`=`	`17`	Equation `B`

Next, multiply the values of equation `A` by `3` and add the product to equation `B` to find the value of `x`.

`3xx``(9x-y)`	`=`	`3xx``7`	Multiply equation `A` by `3`
`27x-3y`	`=`	`21`

`27x-3y`	`=`	`21`
`+` `11x+3y`	`=`	`17`

`38x`	`=`	`38`	`-3y+3y` cancels out
`38x``div38`	`=`	`38``div38`	Divide both sides by `38`
`x`	`=`	`1`

Now, substitute the value of `x` into equation `A` to solve for `y`.

`9``x` `-y`	`=`	`7`	Equation `A`
`9(``1``)-y`	`=`	`7`	`x=1`
`9-y`	`=`	`7`
`9-y` `-9`	`=`	`7` `-9`	Subtract `9` from both sides
`-y`	`=`	`-2`
`-y` `times(-1)`	`=`	`-2` `times(-1)`	Multiply both sides by `-1`
`y`	`=`	`2`

Finally, substitute the value of `x` and `y` into equation `1` to solve for `z`.

`3``x``-``y` `+2z`	`=`	`5`	Equation `1`
`3(``1``)-``2` `+2z`	`=`	`5`	`x=1` and `y=2`
`3-2+2z`	`=`	`5`
`1+2z`	`=`	`5`
`1+2z` `-1`	`=`	`5` `-1`	Subtract `1` from both sides
`2z`	`=`	`4`
`2z` `div2`	`=`	`4` `div2`	Divide both sides by `2`
`z`	`=`	`2`

`x=1`

`y =2`

`z =2`

Question 2 of 5

2. Question

Solve the following systems of equations by elimination.

`3x+5y+z=5`

`9x+2y+7z=32`

`6x+3y+4z=19`

`x=` (3)

`y=` (-1)

`z=` (1)

Hint

Help Video

Correct

Keep Going!

Incorrect

Elimination Method

`1)` make sure 2 equations have a variable with same coefficients but opposite signs
`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 equations `1`, `2` and `3` respectively.

`3x+5y+z`	`=`	`5`	Equation `1`
`9x+2y+7z`	`=`	`32`	Equation `2`
`6x+3y+4z`	`=`	`19`	Equation `3`

Next, multiply the values of equation `1` by `3` and subtract equation `1` from the product. Label the result as equation `A`.

`3xx``(3x+5y+z)`	`=`	`3xx``5`	Multiply equation `1` by `3`
`9x+15y+3z`	`=`	`15`

`9x+15y+3z`	`=`	`15`
`-` `9x-2y-7z`	`=`	`32`	`9x-9x` cancels out

`13y-4z`	`=`	`-17`	Equation `A`

Then, multiply the values of equation `1` by `2` and subtract equation `3` from the product. Label the result as equation `B`.

`2xx``(3x+5y+z)`	`=`	`2xx``5`	Multiply equation `1` by `2`
`6x+10y+2z`	`=`	`10`

`6x+10y+2z`	`=`	`10`
`-` `6x-3y-4z`	`=`	`19`	`6x-6x` cancels out

`7y-2z`	`=`	`-9`	Equation `B`

Now, multiply the values of equation `B` by `2` and subtract equation `A` from the product to find the value of `y`.

`2xx``(7y-2z)`	`=`	`2xx``-9`	Multiply equation `B` by `2`
`14y-4z`	`=`	`-18`

`14y-4z`	`=`	`-18`
`-` `13y+4z`	`=`	`17`	`-4z+4z` cancels out

`y`	`=`	`-1`

Now, substitute the value of `y` into equation `B` to solve for `z`.

`7``y` `-2z`	`=`	`-9`	Equation `B`
`7(``-1``)-2z`	`=`	`-9`	`y=-1`
`-7-2z`	`=`	`-9`
`-7-2z` `+7`	`=`	`-9` `+7`	Add `7` to both sides
`-2z`	`=`	`-2`
`-2z` `div(-2)`	`=`	`-2` `div(-2)`	Divide both sides by `-2`
`z`	`=`	`1`

Finally, substitute the value of `y` and `z` into equation `1` to solve for `x`.

`3x+5``y``+``z`	`=`	`5`	Equation `1`
`3x+5(``-1``)+``1`	`=`	`5`	`y=-1` and `z=1`
`3x-5+1`	`=`	`5`
`3x-4`	`=`	`5`
`3x-4` `+4`	`=`	`5` `+4`	Add `4` to both sides
`3x`	`=`	`9`
`3x` `div3`	`=`	`9` `div3`	Divide both sides by `3`
`x`	`=`	`3`

`x=3`

`y =-1`

`z =1`

Question 3 of 5

3. Question

Solve the following systems of equations by elimination.

`x+2y+5z=2`

`3x+4y-4z=21`

`2x-y+z=3`

`x=` (3)

`y=` (2)

`z=` (-1)

Hint

Help Video

Correct

Excellent!

Incorrect

Elimination Method

`1)` make sure 2 equations have a variable with same coefficients but opposite signs
`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 equations `1`, `2` and `3` respectively.

`x+2y+5z`	`=`	`2`	Equation `1`
`3x+4y-4z`	`=`	`21`	Equation `2`
`2x-y+z`	`=`	`3`	Equation `3`

Next, multiply the values of equation `1` by `3` and subtract equation `2` from the product. Label the result as equation `A`.

`3xx``(x+2y+5z)`	`=`	`3xx``2`	Multiply equation `1` by `3`
`3x+6y+15z`	`=`	`6`

`3x+6y+15z`	`=`	`6`
`-` `3x-4y+4z`	`=`	`21`	`3x-3x` cancels out

`2y+19z`	`=`	`-15`	Equation `A`

Then, multiply the values of equation `1` by `2` and subtract equation `3` from the product. Label the result as equation `B`.

`2xx``(x+2y+5z)`	`=`	`2xx``2`	Multiply equation `1` by `2`
`2x+4y+10z`	`=`	`4`

`2x+4y+10z`	`=`	`4`
`-` `2x+y-z`	`=`	`-3`	`2x-2x` cancels out

`5y+9z`	`=`	`1`	Equation `B`

Now, multiply the values of equation `A` by `5` and the values of equation `B` by `2`, then get their difference to find the value of `z`.

`5xx``(2y+19z)`	`=`	`5xx``-15`	Multiply equation `A` by `5`
`10y+95z`	`=`	`-75`

`2xx``(5y+9z)`	`=`	`2xx``1`	Multiply equation `B` by `2`
`10y+18z`	`=`	`2`

`10y+18z`	`=`	`2`
`-` `10y-95z`	`=`	`75`	`10y-10y` cancels out

`-77z`	`=`	`77`
`-77z` `div(-77)`	`=`	`77` `div(-77)`	Divide both sides by `-77`
`z`	`=`	`-1`

Now, substitute the value of `z` into equation `B` to solve for `y`.

`5y+9``z`	`=`	`1`	Equation `B`
`5y+9(``-1``)`	`=`	`1`	`z=-1`
`5y-9`	`=`	`1`
`5y-9` `+9`	`=`	`1` `+9`	Add `9` to both sides
`5y`	`=`	`10`
`5y` `div5`	`=`	`10` `div5`	Divide both sides by `5`
`y`	`=`	`2`

Finally, substitute the value of `y` and `z` into equation `1` to solve for `x`.

`x+2``y``+5``z`	`=`	`2`	Equation `1`
`x+2(``2``)+5(``-1``)`	`=`	`2`	`y=2` and `z=-1`
`x+4-5`	`=`	`2`
`x-1`	`=`	`2`
`x-1` `+1`	`=`	`2` `+1`	Add `1` to both sides
`x`	`=`	`3`

`x=3`

`y =2`

`z =-1`

Question 4 of 5

4. Question

Solve the following systems of equations by elimination.

`-2x-y+5z=5`

`3x+2y-z=7`

`-4x+4y+2z=12`

`x=` (1)

`y=` (3)

`z=` (2)

Hint

Help Video

Correct

Well Done!

Incorrect

Elimination Method

`1)` make sure 2 equations have a variable with same coefficients but opposite signs
`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 equations `1`, `2` and `3` respectively.

`-2x-y+5z`	`=`	`5`	Equation `1`
`3x+2y-z`	`=`	`7`	Equation `2`
`-4x+4y+2z`	`=`	`12`	Equation `3`

Next, multiply the values of equation `1` by `2` and add equation `2` to the product. Label the result as equation `A`.

`2xx``(-2x-y+5z)`	`=`	`2xx``5`	Multiply equation `1` by `2`
`-4x-2y+10z`	`=`	`10`

`-4x-2y+10z`	`=`	`10`
`+` `3x+2y-z`	`=`	`7`	`-2y+2y` cancels out

`-x+9z`	`=`	`17`	Equation `A`

Then, multiply the values of equation `2` by `2` and subtract equation `3` from the product. Label the result as equation `B`.

`2xx``(3x+2y-z)`	`=`	`2xx``7`	Multiply equation `2` by `2`
`6x+4y-2z`	`=`	`14`

`6x+4y-2z`	`=`	`14`
`-` `4x-4y-2z`	`=`	`-12`	`4y-4y` cancels out

`10x-4z`	`=`	`2`	Equation `B`

Now, multiply the values of equation `A` by `10` and add the product to equation `A` to find the value of `z`.

`10xx``(-x+9z)`	`=`	`10xx``17`	Multiply equation `A` by `10`
`-10x+90z`	`=`	`170`

`10x-4z`	`=`	`2`
`+` `-10x+90z`	`=`	`170`	`10x-10x` cancels out

`86z`	`=`	`172`
`86z` `div86`	`=`	`172` `div86`	Divide both sides by `86`
`z`	`=`	`2`

Now, substitute the value of `z` into equation `B` to solve for `x`.

`10x-4``z`	`=`	`2`	Equation `B`
`10x-4(``2``)`	`=`	`2`	`y=-1`
`10x-8`	`=`	`2`
`10x-8` `+8`	`=`	`2` `+8`	Add `8` to both sides
`10x`	`=`	`10`
`10x` `div10`	`=`	`10` `div10`	Divide both sides by `10`
`x`	`=`	`1`

Finally, substitute the value of `x` and `z` into equation `2` to solve for `y`.

`3``x``+2y-``z`	`=`	`7`	Equation `2`
`3(``1``)+2y-``2`	`=`	`7`	`x=1` and `z=2`
`3+2y-2`	`=`	`7`
`1+2y`	`=`	`7`
`1+2y` `-1`	`=`	`7` `-1`	Subtract `1` from both sides
`2y`	`=`	`6`
`2y` `div2`	`=`	`6` `div2`	Divide both sides by `2`
`y`	`=`	`3`

`x=1`

`y =3`

`z =2`

Question 5 of 5

5. Question

Solve the following systems of equations by elimination.

`2a-b-4c=1`

`a+6b-c=8`

`3a+4b-2c=11`

`a=` (3)

`b=` (1)

`c=` (1)

Hint

Help Video

Correct

Great Work!

Incorrect

Elimination Method

`1)` make sure 2 equations have a variable with same coefficients but opposite signs
`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 equations `1`, `2` and `3` respectively.

`2a-b-4c`	`=`	`1`	Equation `1`
`a+6b-c`	`=`	`8`	Equation `2`
`3a+4b-2c`	`=`	`11`	Equation `3`

Next, multiply the values of equation `2` by `2` and subtract equation `1` from the product. Label the result as equation `A`.

`2xx``(a+6b-c)`	`=`	`2xx``8`	Multiply equation `2` by `2`
`2a+12b-2c`	`=`	`16`

`2a+12b-2c`	`=`	`16`
`-` `2a+b+4c`	`=`	`1`	`2a-2a` cancels out

`13b+2c`	`=`	`15`	Equation `A`

Then, multiply the values of equation `2` by `3` and subtract equation `3` from the product. Label the result as equation `B`.

`3xx``(a+6b-c)`	`=`	`3xx``8`	Multiply equation `2` by `3`
`3a+18b-3c`	`=`	`24`

`3a+18b-3c`	`=`	`24`
`-` `3a-4b+2c`	`=`	`-11`	`6x-6x` cancels out

`14b-c`	`=`	`13`	Equation `B`

Now, multiply the values of equation `B` by `2` and add equation `A` to the product to find the value of `y`.

`2xx``(14b-c)`	`=`	`2xx``13`	Multiply equation `B` by `2`
`28b-2c`	`=`	`26`

`28b-2c`	`=`	`26`
`+` `13b+2c`	`=`	`15`	`-4z+4z` cancels out

`41b`	`=`	`41`
`41b` `div41`	`=`	`41` `div41`	Divide both sides by `41`
`b`	`=`	`1`

Now, substitute the value of `b` into equation `A` to solve for `c`.

`13``b` `+2c`	`=`	`15`	Equation `A`
`13(``1``)+2c`	`=`	`15`	`y=1`
`13+2c`	`=`	`15`
`13+2c` `-13`	`=`	`15` `-13`	Subtract `13` from both sides
`2c`	`=`	`2`
`2c` `div2`	`=`	`2` `div2`	Divide both sides by `2`
`c`	`=`	`1`

Finally, substitute the value of `b` and `c` into equation `2` to solve for `a`.

`a+6``b``-``c`	`=`	`8`	Equation `1`
`a+6(``1``)-``1`	`=`	`8`	`b=1` and `c=1`
`a+6-1`	`=`	`8`
`a+5`	`=`	`8`
`a+5` `-5`	`=`	`8` `-5`	Subtract `5` from both sides
`a`	`=`	`3`

`a=3`

`b =1`

`c =1`

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