How do you solve $2 x - 3 y = 1$ $2x-3y=1$ and $5 x + 4 y = 14$ $5x+4y=14$ using substitution?

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Answer 1 · 2018-06-30T15:59:17

$x = 2, y = 1$ $x=2,y=1$

solving the first equation for $x$ $x$ :
substracting $3 y$ $3y$

$2 x = 1 + 3 y$ $2x=1+3y$
dividing by $2$ $2$ we get
$x = \frac{1}{2} + \frac{3}{2} y$ $x=1/2+3/2y$
plugging this in the second equation

$5 (\frac{1}{2} + \frac{3}{2} y) + 4 y = 14$ $5(1/2+3/2y)+4y=14$

multiplying by $2$ $2$

$5 (1 + 3 y) + 8 y = 28$ $5(1+3y)+8y=28$
$5 + 15 y + 8 y = 28$ $5+15y+8y=28$

collecting like Terms

$23 y + 5 = 28$ $23y+5=28$

substracting $5$ $5$ and dividing by $23$ $23$

$y = 1$ $y=1$

so we get

$x = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$ $x=1/2+3/2=4/2=2$

How do you solve 2x-3y=12x−3y=12x-3y=1 and 5x+4y=145x+4y=145x+4y=14 using substitution?