How do you solve $a + b = 4$ and $2a + 3b = 11$ using substitution?

Question

2972 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-29T14:35:17

$a,b=1,3$

$color(blue)(a+b=4),color(red)(2a+3b=11)$

Solve for first equation

$rarra+b=4$

$rarrb=4-a$

Substitute the value of $b$ to the second equation

$rarr2(4-a)+3b=11$

$rarr8-2b+3b=11$

$rarr8+b=11$

$rarrb=11-8$

$color(green)(rArrb=3$

So substitute the value to the first equation

$rarra+3=4$

$rarra=4-3$

$color(green)(rArra=1$

How do you solve a + b = 4a + b = 4 and 2a + 3b = 112a + 3b = 11 using substitution?