How do you solve $3 a - b = 11$ $3a - b = 11$ and $2 a + 3 b = 0$ $2a + 3b = 0$ using substitution?

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Answer 1 · 2018-03-06T14:14:08

$a = 3, b = - 2$ $color(green)(a = 3, b = -2$

$3 a - b = 11$ $3a - b = 11$ Eqn (1)

$b = 3 a - 11$ $b = 3a - 11$

$2 a + 3 b = 0$ $2a + 3b = 0$ Eqn (2)

Substituting value of b in terms of a in Eqn (2),

$2 a + 9 a - 33 = 0$ $2a + 9a - 33 = 0$

$11 a = 33$ $11a = 33$ or $a = 3$ $color(green)(a = 3$

Substituting value of a in Eqn (1),

$(3 \cdot 3) - b = 11$ $(3 * 3) - b = 11$

$b = 9 - 11$ $b = 9 - 11$ , $b = - 2$ $color(green)( b = - 2$

How do you solve 3a - b = 113a−b=113a - b = 11 and 2a + 3b = 02a+3b=02a + 3b = 0 using substitution?