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How do I solve this system of equations?

$$\begin{cases} 7(a+b)=b-a \\4(3a+2b)=b-8\end{cases}$$

Progress

I tried both substitution and elimination, but when I set $a$ or $b$ free on one side, I keep getting $a$ or $b$ also on the other side.

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  • $\begingroup$ I tried both substitution and elimination, but when I set A or B free on one side, I keep getting A or B also on the other side. $\endgroup$ – martjah Sep 6 '14 at 14:48
  • $\begingroup$ Start by collecting all the unknowns on one side of each equation, constants on the other side. $\endgroup$ – hardmath Sep 6 '14 at 14:51
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    $\begingroup$ Learn about the canonical method to solve this kind of system: Cramer's rule. en.wikipedia.org/wiki/… $\endgroup$ – Yves Daoust Sep 6 '14 at 15:18
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$$7 \cdot (a+b) = b - a \Rightarrow a = - \dfrac{6}{8} b$$

Substitute for a in the second equation

$$4 \cdot (3a+ 2b) = b-8 \Rightarrow 12a + 7b = -8$$

$$7b + 12 \cdot \left( - \dfrac{6}{8}b \right) = -8$$

This equation has only $b$ unknown so solve for $b$ then use $b$ to find $a$

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Hint:

Try to perform algebra on both the equations till you have the $a$ and $b$ on one side and a number on the other. Then see if you can "combine" them together.

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$$7(a+b) = b-a \Rightarrow 7a+7b=b-a \Rightarrow 7a+a=b-7b \Rightarrow 8a=-6b \Rightarrow a=-\frac{6}{8}b \\ \Rightarrow a=-\frac{3}{4}b\\ 4(3a+2b)=b-8 \Rightarrow 12a+8b=b-8 \Rightarrow 12a=b-8b-8 \Rightarrow 12a=-7b-8 \overset{a=-\frac{3}{4}b}{\Rightarrow} 12 \left (-\frac{3}{4}b \right ) =-7b-8 \Rightarrow -9b=-7b-8 \Rightarrow -9b+7b=-8 \Rightarrow -2b=-8 \\ \Rightarrow b=4$$

Replacing at $a=-\frac{3}{4}b$ we get $a=-\frac{3}{4}4 \Rightarrow a=-3$

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CAUTION: this is a non-standard approach.

In both equations, isolate the variable $a$ in the LHS: $$7(a+b)=b-a\implies 8a=-6b\implies 4a=-3b,$$ $$4(3a+2b)=b-8\implies12a=-7b-8.$$ Now equate the two: $$(12a=)-9b=-7b-8.$$ This is an equation in a single unknown ($b$). $$-2b=-8\implies b=4,$$ and $$4a=-3b\implies a=-3.$$

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Hint: from the first equation, we have $$a=-\frac{3}{4}b.$$

Now substitute this into the second equation and you've now got an equation in $b$. Solve this (for $b$), then find $a$ (again) using the fact that $$a=-\frac{3}{4}b.$$

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Here is a different approach:

$\begin{cases} 7(a+b)=b-a \\ 4(3a+2b) = b-8 \end{cases}$ $\Leftrightarrow$ $\begin{cases}8a +6b=0\\12a+7b=-8 \end{cases}$

Then

$\begin{bmatrix} 8&6&0\\12&7&-8\end{bmatrix}$ $\Leftrightarrow$ $\begin{bmatrix} 1&3/4&0\\12&7&-8 \end{bmatrix}$ $\Leftrightarrow$ $\begin{bmatrix}1&3/4&0\\0&-2&-8\end{bmatrix}$ $\Leftrightarrow$ $\begin{bmatrix}1&3/4&0\\0&1&4\end{bmatrix}$ $\Leftrightarrow$ $\begin{bmatrix}1&0&-3\\0&1&4\end{bmatrix}$

Resulting in $a=-3$ aand $b=4$.

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