Rewriting Equations

**vaironxxrd** · September 14th 2011, 03:23 PM

I have to solve for y . No given values
I already tried solving this but I just can't solve the fraction. I tried multiplying the reciprocal on both sides but I don't know if I have to make 20 a fraction
$-\frac{5}{4}&x+&5y&=20$

**pickslides** · September 14th 2011, 03:34 PM

$-\frac{5}{4}x+5y=20$

$5y=20+\frac{5}{4}x$

$y=\frac{20+\frac{5}{4}x}{5}$

$y=4+\frac{x}{4}$

**vaironxxrd** · September 14th 2011, 05:30 PM

Originally Posted by pickslides

$-\frac{5}{4}x+5y=20$

$5y=20+\frac{5}{4}x$

$y=\frac{20+\frac{5}{4}x}{5}$

$y=4+\frac{x}{4}$

Whats going on with the fraction in the second step? is it being simplified by 5?

**skeeter** · September 14th 2011, 05:35 PM

$-\frac{5}{4}x+5y=20$

$5y=20+\frac{5}{4}x$

multiply every term by 4 to clear the fraction ...

$20y=80+5x$

divide every term by by 20 ...

$y = 4 + \frac{x}{4}$

**vaironxxrd** · September 15th 2011, 12:06 PM

Originally Posted by skeeter

$-\frac{5}{4}x+5y=20$

$5y=20+\frac{5}{4}x$

multiply every term by 4 to clear the fraction ...

$20y=80+5x$

divide every term by by 20 ...

$y = 4 + \frac{x}{4}$

I got everything until you have to divide by 20.

80/20 = 4
5x/20 = Lost

**e^(i*pi)** · September 15th 2011, 12:07 PM

Originally Posted by vaironxxrd

I got everything until you have to divide by 20.

80/20 = 4
5x/20 = Lost

$\frac{5x}{20} = \dfrac{5x}{5 \times 4}$

Cancel 5 to get x/4

**vaironxxrd** · September 15th 2011, 12:10 PM

Originally Posted by e^(i*pi)

$\frac{5x}{20} = \dfrac{5x}{5 \times 4}$

Cancel 5 to get x/4

Got it , I din't know I could do that,
Thanks

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