How do you write $y = \frac{3}{8} x - \frac{3}{4}$ $y=3/8x-3/4$ in standard form?

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Answer 1 · 2015-06-28T06:50:55

$y = \frac{3}{8} x - \frac{3}{4}$ $y=3/8x-3/4$ in standard form is $3 x - 8 y = 6$ $3x-8y=6$ .

Standard form for a linear equation is $A x + B y = C$ $Ax+By=C$ .

$y = \frac{3}{8} x - \frac{3}{4}$ $y=3/8x-3/4$

Simplify $\frac{3}{8} x$ $3/8x$ to $\frac{3 x}{8}$ $(3x)/8$ .

Subtract $\frac{3 x}{8}$ $(3x)/8$ from both sides of the equation.

$\frac{- 3 x}{8} + y = - \frac{3}{4}$ $(-3x)/8+y=-3/4$

Multiply both sides by $- 1$ $-1$ .

$\frac{- 3 x}{8} (- 1) + y (- 1) = - \frac{3}{4} (- 1)$ $(-3x)/8(-1)+y(-1)=-3/4(-1)$ =

$\frac{3 x}{8} - y = \frac{3}{4}$ $(3x)/8-y=3/4$

Multiply both sides by $8$ $8$ .

$\frac{(3 x) \cdot 8}{8} - y (8) = \frac{(3) \cdot 8}{4}$ $((3x)*8)/8-y(8)=((3)*8)/4$ =

$((3x)*cancel8^1)/(cancel 8^1)-y(8)=((3)*cancel8^2)/(cancel4^1)$ =

$3x-8y=6$

How do you write y=3/8x-3/4y=38x−34y=3/8x-3/4 in standard form?