Respuesta :

[tex]\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}[/tex]

[tex]\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20[/tex]
[tex]\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12[/tex]

[tex]\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}[/tex]

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

[tex]\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix} [/tex]

[tex]3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides} [/tex]
[tex]3a-10y-3a=-8-3a[/tex]

[tex]\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10[/tex]
[tex]\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}[/tex]

Simplify more.

[tex]\frac{-10y}{-10} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{10y}{10} [/tex]

[tex]\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \ y[/tex]

[tex]-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{-8-3a}{-10}[/tex]

[tex]\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-3a-8}{10} \ \textgreater \ y=-\frac{-8-3a}{10}[/tex]

[tex]\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \ 6x+10\cdot \frac{8}{10-3a}=20[/tex]

[tex]10\cdot \frac{8}{10-3a} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{8\cdot \:10}{10-3a}[/tex]
[tex]\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ \frac{80}{10-3a}[/tex]

[tex]6x+\frac{80}{10-3a}=20 \ \textgreater \ \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides} [/tex]
[tex]6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}[/tex]

[tex]\mathrm{Simplify} \ \textgreater \ 6x=20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}[/tex]

[tex]\frac{6x}{6} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \ x[/tex]

[tex]\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{20-\frac{80}{-3a+10}}{6}[/tex]

[tex]20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \ \frac{20}{1}-\frac{80}{-3a+10}[/tex]

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

[tex]Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \ \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}[/tex]

[tex]\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}[/tex]
[tex]\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \ \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}[/tex]

[tex]20\left(-3a+10\right)-80 \ \textgreater \ Rewrite \ \textgreater \ 20+10-3a-4\cdot \:20[/tex]

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

[tex]10-3a-4 \ \textgreater \ \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \ -3a+6 \ \textgreater \ Rewrite[/tex]
[tex]-3a+2\cdot \:3[/tex]

[tex]\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \ 3\left(-a+2\right) \ \textgreater \ 3\cdot \:20\left(-a+2\right) \ \textgreater \ Refine[/tex]
[tex]60\left(-a+2\right)[/tex]

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

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

[tex]Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}[/tex]

Hope this helps!
--------------------------
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
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