How do you solve the following linear system 4x - 3y = -5 , 4x +- 4y = -4 ?

Aug 31, 2017

I have started you off

Explanation:

$4 x - 3 y = - 5 \text{ "->" "y=4/3x +5/3" } \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

$4 x + 4 y = - 4 \text{ "->" "y=-x-1" } \ldots \ldots \ldots \ldots E q u a t i o n \left(2\right)$

$4 x - 4 y = - 4 \text{ "->" "y=x+1" } \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(3\right)$
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Using $E q u a t i o n \left(2\right)$ substitute for $y$ in $E q u a t i o n \left(1\right)$

$y = - x - 1 \text{ "->" } \frac{4}{3} x + \frac{5}{3} = - x - 1$

color(white)("dddddddddddd")->" "7/3x=- 8/3

color(white)("dddddddddddd")->" "x=-8/7

Thus $y = - x - 1 \text{ "->" } y = - \left(- \frac{8}{7}\right) - 1 = + \frac{1}{7}$

So intersection of $E q u a t i o n \left(1\right) \mathmr{and} E q u a t i o n \left(2\right) \to \left(x , y\right) = \left(- \frac{8}{7} , + \frac{1}{7}\right)$
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You repeat this approach for

$E q n \left(2\right) \to E q n \left(3\right)$

Then for

$E q n \left(1\right) \to E q n \left(3\right)$

I will let you do those.
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