Question

For what value of a should you solve the system of elimination?

3x + 5y = 10
2x + ay = 4

Answer 1

--------------------------
3x + 5y = 10:

1. Subtract 5y from both sides

3x = 10 - 5y

2. Divide both sides by 3

x = 10 - 5y over 3

3. Factor out the common term 5

x = 5(2 - y) over 3

Final Answer:

x = 5(2 - y) over 3
-----------------------------
2x + ay = 4

1. Subtract ay from both sides

2x = 4 - ay

2. Divide both sides by 2

x = 4 - ay over 2

Final Answer:

x = 4 - ay over 2

Answer 2

`\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}`


`\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20`

`\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12`


`\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}`

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8


`\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}`


`3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}`

`3a-10y-3a=-8-3a`


`\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10`

`(-10y)/(-10)=-(8)/(-10)-(3a)/(-10)`

Simplify more.


`(-10y)/(-10) \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: (-a)/(-b)=(a)/(b) \ \textgreater \ (10y)/(10)`


`\mathrm{Divide\:the\:numbers:}\:(10)/(10)=1 \ \textgreater \ y`


`-(8)/(-10)-(3a)/(-10) \ \textgreater \ \mathrm{Apply\:rule}\:(a)/(c)\pm (b)/(c)=(a\pm \:b)/(c) \ \textgreater \ (-8-3a)/(-10)`


`\mathrm{Apply\:the\:fraction\:rule}: (a)/(-b)=-(a)/(b) \ \textgreater \ -(-3a-8)/(10) \ \textgreater \ y=-(-8-3a)/(10)`


`\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=(8)/(10-3a) \ \textgreater \ 6x+10\cdot (8)/(10-3a)=20`


`10\cdot (8)/(10-3a) \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c) \ \textgreater \ (8\cdot \:10)/(10-3a)`

`\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ (80)/(10-3a)`


`6x+(80)/(10-3a)=20 \ \textgreater \ \mathrm{Subtract\:}(80)/(10-3a)\mathrm{\:from\:both\:sides}`

`6x+(80)/(10-3a)-(80)/(10-3a)=20-(80)/(10-3a)`


`\mathrm{Simplify} \ \textgreater \ 6x=20-(80)/(10-3a) \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ (6x)/(6)=(20)/(6)-((80)/(10-3a))/(6)`


`(6x)/(6) \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:(6)/(6)=1 \ \textgreater \ x`


`(20)/(6)-((80)/(10-3a))/(6) \ \textgreater \ \mathrm{Apply\:rule}\:(a)/(c)\pm (b)/(c)=(a\pm \:b)/(c) \ \textgreater \ (20-(80)/(-3a+10))/(6)`


`20-(80)/(10-3a) \ \textgreater \ \mathrm{Convert\:element\:to\:fraction}: \:20=(20)/(1) \ \textgreater \ (20)/(1)-(80)/(-3a+10)`


`\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10`


`Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \ (20\left(-3a+10\right))/(-3a+10)-(80)/(-3a+10)`


`\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: (a)/(c)\pm (b)/(c)=(a\pm \:b)/(c)`

`(20\left(-3a+10\right)-80)/(-3a+10) \ \textgreater \ ((20\left(-3a+10\right)-80)/(-3a+10))/(6) \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: ((b)/(c))/(a)=(b)/(c\:\cdot \:a)`


`20\left(-3a+10\right)-80 \ \textgreater \ Rewrite \ \textgreater \ 20+10-3a-4\cdot \:20`


`\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \ 20\left(-3a+10-4\right) \ \textgreater \ Factor\;more`


`10-3a-4 \ \textgreater \ \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \ -3a+6 \ \textgreater \ Rewrite`

`-3a+2\cdot \:3`


`\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \ 3\left(-a+2\right) \ \textgreater \ 3\cdot \:20\left(-a+2\right) \ \textgreater \ Refine`

`60\left(-a+2\right)`


`(60\left(-a+2\right))/(6\left(-3a+10\right)) \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:(60)/(6)=10 \ \textgreater \ (10\left(-a+2\right))/(\left(-3a+10\right))`


`\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \ (10\left(-a+2\right))/(-3a+10)`


`Therefore\;our\;solutions\;are\; y=(8)/(10-3a),\:x=(10\left(-a+2\right))/(-3a+10)`

Hope this helps!