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

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Answer 1 · 2016-02-29T14:35:17

$a, b = 1, 3$ $a,b=1,3$

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

Solve for first equation

$\to a + b = 4$ $rarra+b=4$

$\to b = 4 - a$ $rarrb=4-a$

Substitute the value of $b$ $b$ to the second equation

$\to 2 (4 - a) + 3 b = 11$ $rarr2(4-a)+3b=11$

$\to 8 - 2 b + 3 b = 11$ $rarr8-2b+3b=11$

$\to 8 + b = 11$ $rarr8+b=11$

$\to b = 11 - 8$ $rarrb=11-8$

$\Rightarrow b = 3$ $color(green)(rArrb=3$

So substitute the value to the first equation

$\to a + 3 = 4$ $rarra+3=4$

$\to a = 4 - 3$ $rarra=4-3$

$\Rightarrow a = 1$ $color(green)(rArra=1$

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