Question

Solve the following system of equations by substitution and select the correct answer below: 6x − 4y = 36 2x − 8y = 32 A. x=4, y=3 B. x=4, y=-3 C. x=-4, y=3 D. x=-4, y=-3

Answer 1

Via subsitution we want to isolate x for 6x - 4y = 36
Start by adding 4y to both sides.

`6x-4y+4y=36+4y \ \textgreater \ Simplify \ \textgreater \ 6x=36+4y`
Now divide both sides by 6

`(6x)/(6)=(36)/(6)+(4y)/(6)`.
Now simplify further.

`(6x)/(6) = (6)/(6) = 1 = x`
Apply the rule of
`(a)/(c)\pm (b)/(c)=(a\pm \:b)/(c)` to
`(36)/(6)+(4y)/(6)`
When applied >
`(36+4y)/(6)` > Now factor 36 + 4y > Rewrite it as....

`4\cdot \:9+4y` > Factor out the common term of 4 >
`4\left(y+9\right)`
This gives us
`(4\left(y+9\right))/(6)`
From there we want to cancel the common factor of 2 which gives us...

`(2\left(y+9\right))/(3)`
Now you want to
`\mathrm{Subsititute\:}x=(2\left(y+9\right))/(3) \ \textgreater \ \begin{bmatrix}2\cdot (2\left(y+9\right))/(3)-8y=32\end{bmatrix}`

`2\cdot (2\left(y+9\right))/(3) \ \textgreater \ \mathrm{Multiply\:fractions}:\quad \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c) \ \textgreater \ (2\left(y+9\right) \cdot \:2)/(3)`

`\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4`
This gives us
`(4\left(y+9\right))/(3)`
So now we have
`(4\left(y+9\right))/(3)-8y=32`
We want to multiply both sides by 3.

`(4\left(y+9\right))/(3)\cdot \:3-8y\cdot \:3=32\cdot \:3`
Refine it.

`4\left(y+9\right)-24y=96`
Expand
`4\left(y+9\right)`

`\mathrm{Distribute\:parentheses\:using}:\quad \:a\left(b+c\right)=ab+ac`
Where
`a=4,\:b=y,\:c=9`

`4\cdot \:y+4\cdot \:9 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:4\cdot \:9=36 \ \textgreater \ 4y +36`
So now we have
`4y+36-24y`
Group the like terms.

`4y-24y+36 \ \textgreater \ \mathrm{Add\:similar\:elements:}\:4y-24y=-20y \ \textgreater \ -20y+36`

`-20y+36=96 \ \textgreater \ \mathrm{Subtract\:}36\mathrm{\:from\:both\:sides}`

`-20y+36-36=96-36 \ \textgreater \ Simplify \ \textgreater \ -20y=60 \ \textgreater \`
Now we want to
`\mathrm{Divide\:both\:sides\:by\:}-20 \ \textgreater \ (-20y)/(-20)=(60)/(-20) \ \textgreater \ y=-3`

`\mathrm{For\:}x=(2\left(y+9\right))/(3) \ \textgreater \ \mathrm{Subsititute\:}y=-3 \ \textgreater \ x=(2\left(-3+9\right))/(3)\quad \Rightarrow \quad x=4`
Therefore our final solutions are, y = -3, x =4.
Hope this helps!