How do you solve $7 a = 3 - 5 b$ $7a=3-5b$ and $2 a + b = 0$ $2a+b=0$ ?

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Answer 1 · 2016-05-09T15:04:29

The solution for the system of equations is:
$a = - 1$ $color(green)(a = -1$

$b = 2$ $color(green)(b = 2$

$7 a = 3 - 5 b$ $7a = 3 - 5b$ ..........equation $(1)$ $(1)$

$2 a + b = 0$ $2a + b = 0$

$b = - 2 a$ $color(green)(b = -2a$ ..............equation $(2)$ $(2)$

Solving by substitution:

Substituting equation $2$ $2$ in $1$ $1$ :

$7 a = 3 - 5 b$ $7a = 3 - 5b$

$7 a = 3 + (- 5) \cdot (- 2 a)$ $7a = 3 + (- 5) * color(green)((-2a))$

$7 a = 3 + 10 a$ $7a = 3 + 10a$

$7 a - 10 a = 3$ $7a - 10a = 3$

$- 3 a = 3$ $- 3a = 3$

$a = \frac{3}{- 3}$ $a = 3 / (-3)$

$a = - 1$ $color(green)(a = -1$

Finding $b$ $b$ from equation $2$ $2$ :

$b = - 2 a$ $b = -2a$

$b = - 2 \cdot (- 1)$ $b = -2 * (-1)$

$b = 2$ $color(green)(b = 2$

How do you solve 7a=3-5b 7a=3−5b7a=3-5b and 2a+b=02a+b=02a+b=0?