Problem 6

Solve for $x$ in terms of $a$, $b$, and $c$ from

$\frac{a-2x}{b}=\frac{cx-bc}{a}$

and then find $x$ when $a=2$, $b=-1$, $c=3$

Solution

a) Multiply both sides by $ab$ assuming $ab\ne 0$: $a(a-2x)=b(cx-bc)$

b) Expand both sides: $a^2-2ax=bcx-b^{2}c$

c) Collect the $x$ terms: $a^2+b^{2}c=x(bc+2a)$

d) Solve for $x$: $x=\frac{a^2+b^{2}c}{2a+bc}$

e) For $a=2$, $b=-1$, $c=3$:

The numerator is

$2^2+{(-1)}^2\cdot 3=4+3=7$

and the denominator is

$2\left(2\right)+(-1)\left(3\right)=4-3=1.$

Hence, $x=7.$

Answer: $x=\frac{a^2+b^2с}{2a+bc}$ and for $a=2;b=-1;c=3;x=7.$