Question

Can you solve this?????
8x-6y=20 and -16x+7y=30: Answer this please please!

Answer 1

`\begin{bmatrix}8x-6y=20\\ -16x+7y=30\end{bmatrix}`


`Isolate\;x\;for\;8x-6y=20 \ \textgreater \ \mathrm{Add\:}6y\mathrm{\:to\:both\:sides}`

`8x-6y+6y=20+6y \ \textgreater \ \mathrm{Simplify} \ \textgreater \ 8x=20+6y`

`\mathrm{Divide\:both\:sides\:by\:}8 \ \textgreater \ (8x)/(8)=(20)/(8)+(6y)/(8) \ \textgreater \ Simplify`


`(8x)/(8) \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:(8)/(8)=1 \ \textgreater \ x`


`(20)/(8)+(6y)/(8) \ \textgreater \ \mathrm{Apply\:rule}\:(a)/(c)\pm (b)/(c)=(a\pm \:b)/(c) \ \textgreater \ (20+6y)/(8) \ \textgreater \ Factor`

`20+6y \ \textgreater \ Rewrite \ \textgreater \ 2\cdot \:10+2\cdot \:3y \ \textgreater \ \mathrm{Factor\:out\:common\:term\:}2`

`2\left(3y+10\right) \ \textgreater \ (2\left(3y+10\right))/(8) \ \textgreater \ \mathrm{Cancel\:the\:common\:factor:}\:2 \ \textgreater \ x = (3y+10)/(4)`

So now..

`\mathrm{Subsititute\:}x=(3y+10)/(4) \ \textgreater \ \begin{bmatrix}-16\cdot (3y+10)/(4)+7y=30\end{bmatrix}`


`Isolate\;y\;for\;-16\cdot (3y+10)/(4)+7y=30`


`16\cdot (3y+10)/(4) \ \textgreater \ \mathrm{Multiply\:fractions}:\ \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c) \ \textgreater \ (\left(3y+10\right)\cdot \:16)/(4)`

`\mathrm{Divide\:the\:numbers:}\:(16)/(4)=4 \ \textgreater \ 4\left(3y+10\right) \ \textgreater \ -4\left(3y+10\right)+7y=30`


`Expand\;-4\left(3y+10\right)+7y \ \textgreater \ -4\left(3y+10\right)`


`\mathrm{Distribute\:parentheses\:using}: \:a\left(b+c\right)=ab+ac`

`Where\;a=-4,\:b=3y,\:c=10 \ \textgreater \ -4\cdot \:3y-4\cdot \:10`


`-4\cdot \:3y-4\cdot \:10 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:4\cdot \:3=12 \ \textgreater \ -12y-4\cdot \:10`

`\mathrm{Multiply\:the\:numbers:}\:4\cdot \:10=40 \ \textgreater \ -12y-40 \ \textgreater \ -12y-40+7y`


`Simplify\;-12y-40+7y \ \textgreater \ \mathrm{Group\:like\:terms} \ \textgreater \ -12y+7y-40`

`\mathrm{Add\:similar\:elements:}\:-12y+7y=-5y \ \textgreater \ -5y-40 \ \textgreater \ -5y-40 = 30`


`\mathrm{Add\:}40\mathrm{\:to\:both\:sides} \ \textgreater \ -5y-40+40=30+40 \ \textgreater \ -5y=70`


`\mathrm{Divide\:both\:sides\:by\:}-5 \ \textgreater \ (-5y)/(-5)=(70)/(-5) \ \textgreater \ Simplify \ \textgreater \ y=-14`


`\mathrm{For\:}x=(3y+10)/(4) \ \textgreater \ \mathrm{Subsititute\:}y=-14 \ \textgreater \ x=(3\left(-14\right)+10)/(4)\quad \Rightarrow \quad x=-8`


`Therefore\;the\:solutions\;are\;\ \textgreater \ y=-14,\:x=-8`

Hope this helps!