What is 5x+4y-4=0,x-20=12y ?

The solution to 5x+4y-4=0,x-20=12y is x=2,y=-3/2

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5x+4y-4=0,x-20=12y

5 x + 4 y - 4 = 0, x - 20 = 12 y

x = 2, y = - \frac{3}{2}

Solution steps

[5 x + 4 y - 4 = 0 x - 20 = 12 y]

Isolate

x

for

5 x + 4 y - 4 = 0 : x = \frac{- 4 y + 4}{5}

Substitute

x = \frac{- 4 y + 4}{5}

[\frac{- 4 y + 4}{5} - 20 = 12 y]

Isolate

y

for

\frac{- 4 y + 4}{5} - 20 = 12 y : y = - \frac{3}{2}

For

x = \frac{- 4 y + 4}{5}

Substitute

y = - \frac{3}{2}

x = \frac{- 4 ( - \frac{3}{2} ) + 4}{5}

\frac{- 4 ( - \frac{3}{2} ) + 4}{5} = 2

x = 2

The solutions to the system of equations are:

x = 2, y = - \frac{3}{2}

c t^{2} = 80, c v^{1} = 625 l v^{2}, t^{1} = 0

f (x) = x^{2} + 2 x, x^{0} = 1.5

7 x + 5 y = 13, 3 x - 2 y = - 11

- x + 3 y = 8, 4 x + 7 y = 25

a + 2 b = 3, 2 a + b = 0