Question

Solve the following system of equations for a and for b:

Answer 1

Answer:

$a=3\\b=1$

Step-by-step explanation:

$9a+3b=30\\8a+4b=28$

Let's solve the second equation for a to later on replace it in the first equation.

$8a+4b=28\\8a=28-4b\\a=\frac{28-4b}{8}$

Now plug this into the first equation.

$9a+3b=30\\9(\frac{28-4b}{8})+3b=30$

Distribute the 9

$(\frac{252-36b}{8}) +3b=30$

Break down the fraction.

$\frac{252}{8}-\frac{36b}{8}+3b=30$

Simplify.

$\frac{63}{2}-\frac{9}{2}b+3b=30$

Subtract $\frac{63}{2}$

$-\frac{9}{2}b+3b=30-\frac{63}{2}$

Combine like terms.

$\frac{-9+2*3}{2}b=\frac{30*2-63}{2}$

$\frac{-9+6}{2}b=\frac{60-63}{2}$

$\frac{-3}{2}b=\frac{-3}{2}$

Muliply by the reciprocal or inverted fraction next to b.

$(-\frac{2}{3})(-\frac{3}{2}) b=-\frac{3}{2}(-\frac{2}{3})$

$b=1$

Now plug this value into any of the equations to find the value of a.

$8a+4b=28\\8a+4(1)=28\\8a+4=28\\8a=28-4\\8a=24\\a=\frac{24}{8}\\ a=3$