How do you solve the system $\frac{x}{3} + \frac{y}{5} = 2$ $x/3+y/5=2$ and $\frac{x}{3} - \frac{y}{2} = - \frac{1}{3}$ $x/3 - y/2= -1/3$ ?

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Answer 1 · 2016-03-25T22:22:08

I did this in a hurry so you need to check the calculations!
$y = \frac{40}{21}$ $y=40/21$

$x = \frac{51}{4}$ $x=51/4$

My first reaction was to consider getting rid of the fractions. That was, until I spotted the coefficients of $x$ $x$ are the same in both equation.

Given:

$\frac{x}{3} + \frac{y}{5} = 2$ $x/3+y/5=2$ ..............................(1)

$\frac{x}{3} - \frac{y}{2} = - \frac{1}{3}$ $x/3-y/2=-1/3$ ...........................(2)

Subtract equation (2) from (1)

$0 + \frac{y}{5} + \frac{y}{2} = 2 + \frac{1}{3}$ $0+y/5+y/2=2+1/3$

$\frac{2 y + 5 y}{10} = \frac{4}{3}$ $(2y+5y)/10=4/3$

$7 y = \frac{40}{3}$ $7y=40/3$

$y = \frac{40}{21}$ $y=40/21$

Now all you have to do is substitute

Substitute in equation (1) giving

$\frac{x}{3} + (\frac{1}{5} \times \frac{40}{21}) = 2$ $x/3+(1/5xx40/21)=2$ .........................(1_a)

$\frac{x}{3} = 2 - \frac{8}{21} = \frac{34}{8}$ $x/3 = 2-8/21 = 34/8$

$x = \frac{3 \times 34}{8} = 12 \frac{3}{4} = \frac{51}{4}$ $x=(3xx34)/8 =12 3/4 =51/4$

How do you solve the system x3+y5=2x/3+y/5=2 and x3−y2=−13x/3 - y/2= -1/3?