How do you solve $3 a - 2 b = 8$ $3a - 2b = 8$ and $4 a + b = 7$ $4a + b = 7$ using substitution?

Question

2589 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-04T11:12:37

From the second equation you may infer: $b = 7 - 4 a$ $b=7-4a$

Now substitute $b$ $b$ in the first equation:

$3 a - 2 (7 - 4 a) = 8 \to 3 a - 14 + 8 a = 8 \to$ $3a-2(7-4a)=8->3a-14+8a=8->$

$3 a + 8 a = 8 + 14 \to 11 a = 22 \to a = \frac{22}{11} = 2$ $3a+8a=8+14->11a=22->a=22/11=2$

$b = 7 - 4 \cdot 2 = - 1$ $b=7-4*2=-1$

Check your answer:
$3 \cdot 2 - 2 \cdot (- 1) = 6 + 2 = 8$ $3*2-2*(-1)=6+2=8$ check!
$4 \cdot 2 + (- 1) = 8 - 1 = 7$ $4*2+(-1)=8-1=7$ check!

How do you solve 3a−2b=83a - 2b = 8 and 4a+b=74a + b = 7 using substitution?