# How do solve the following linear system?: 5x + y = 9 , 3x+ 2y - 3 = 0 ?

May 28, 2018

$\left(x , y\right) \to \left(\frac{15}{7} , - \frac{12}{7}\right)$

#### Explanation:

$5 x + y = 9 \to \left(1\right)$

$3 x + 2 y - 3 = 0 \to \left(2\right)$

$\text{rearrange "(1)" expressing y in terms of x}$

$\text{subtract "5x" from both sides}$

$y = 9 - 5 x \to \left(3\right)$

$\text{substitute "y=9-5x" into equation } \left(2\right)$

$3 x + 2 \left(9 - 5 x\right) - 3 = 0$

$3 x + 18 - 10 x - 3 = 0$

$- 7 x + 15 = 0$

$\text{subtract 15 from both sides}$

$- 7 x = - 15$

$\text{divide both sides by } - 7$

$\frac{\cancel{- 7} x}{\cancel{- 7}} = \frac{- 15}{- 7} \Rightarrow x = \frac{15}{7}$

$\text{substitute "x=15/7" in equation } \left(3\right)$

$y = 9 - \frac{75}{7} = \frac{63}{7} - \frac{75}{7} = - \frac{12}{7}$

$\text{point of intersection } = \left(\frac{15}{7} , - \frac{12}{7}\right)$