# How do you solve 3/4x + 1/3y = 1 and x - y = 10?

Jul 14, 2018

$x = 10 + y$ substitute this into the first equation

$\frac{3}{4} \left(10 + y\right) + \frac{1}{3} y = 1$

$\frac{30}{4} + \frac{3}{4} y + \frac{1}{3} y = 1$

Multiply by 12 to remove the fractions

$90 + 9 y + 4 y = 12$

$90 + 13 y = 12$

$13 y = - 78$

$y = - 6$

Put #y=-6 into equation 2

$x - - 6 = 10$

$x = 4$

Jul 14, 2018

$x = 4 , y = - 6$

#### Explanation:

$\therefore \frac{3}{4} x + \frac{1}{3} y = 1 - - - \left(1\right)$

$\therefore x - y = 10 - - - - \left(2\right)$

$\therefore \left(2\right) \times \frac{3}{4}$

$\therefore \frac{3}{4} x - \frac{3}{4} y = \frac{15}{2} - - - - \left(3\right)$

$\therefore \left(1\right) - \left(3\right)$

$\therefore \frac{13}{12} y = - \frac{13}{2}$

$\therefore y = - \frac{13}{2} \div \frac{13}{12}$

$\therefore y = - {\cancel{13}}^{1} / {\cancel{2}}^{1} \times {\cancel{12}}^{6} / {\cancel{13}}^{1}$

$\therefore y = - 6$

$\text{substitute}$ y=-6$\text{ in} \left(2\right)$

$\therefore x - \left(- 6\right) = 10$

$\therefore x + 6 = 10$

$\therefore x = 10 - 6$

$\therefore x = 4$

~~~~~~~~

check:-

$\text{substitute}$ y=-6$\mathmr{and}$x=3 $\text{in} \left(1\right)$

$\therefore \frac{3}{4} \left(4\right) + \frac{1}{3} \left(- 6\right) = 1$

$\therefore 3 - 2 = 1$

$\therefore 1 = 1$