How do you solve $x + y = 4$ $x+y = 4$ and $5 x - 2 y = - 1$ $5x-2y = -1$ ?

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Answer 1 · 2015-06-01T20:58:34

Starting with $x + y = 4$ $x+y=4$ , subtract $x$ $x$ from both sides to get:

$y = 4 - x$ $y = 4-x$

Then substitute this in the second equation as follows:

$- 1 = 5 x - 2 y$ $-1 = 5x-2y$

$= 5 x - 2 (4 - x) = 5 x - 8 + 2 x = 7 x - 8$ $= 5x-2(4-x) = 5x-8+2x = 7x-8$

Add $8$ $8$ to both ends to get:

$7 x = 7$ $7x = 7$

Divide both sides by $7$ $7$ to get

$x = 1$ $x = 1$

Substitute that value for $x$ $x$ into $y = 4 - x$ $y = 4-x$ to get

$y = 4 - x = 4 - 1 = 3$ $y=4-x=4-1=3$

How do you solve x+y=4x+y = 4 and 5x−2y=−15x-2y = -1?