How do you solve the following system: 8x-3y=3, 4x + 2y = 10 ?

Apr 5, 2016

$\left(x , y\right) = \left(\frac{9}{7} , \frac{17}{7}\right)$

Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} 8 x - 3 y = 3$
$\textcolor{w h i t e}{\text{XXX}} 4 x + 2 y = 10$

If we note that the $x$ term in  is a simple multiple of that in ,
we have an easy way to eliminate the $x$ term:
Multiply  by $2$ and subtract the result from 
$\textcolor{w h i t e}{\text{XXXXXXxX}} 8 x - 3 y = 3$
-$\times 2 \textcolor{w h i t e}{\text{XX}} - \left(\underline{8 x + 4 y = 20}\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXX}} - 7 y = - 17$

Dividing both sides by $\left(- 7\right)$
$\textcolor{w h i t e}{\text{XXX}} y = \frac{17}{7}$

Substituting $\left(\frac{17}{7}\right)$ for $y$ in 
$\textcolor{w h i t e}{\text{XXX}} 4 x + \frac{34}{7} = \frac{70}{7}$

$\textcolor{w h i t e}{\text{XXX}} 4 x = \frac{36}{7}$

$\textcolor{w h i t e}{\text{XXX}} x = \frac{9}{7}$

Apr 5, 2016

$\left(x , y\right) = \left(\frac{9}{7} , \frac{17}{7}\right)$

Explanation:

Solve by elimination and substitution

color(blue)(8x-3y=3

color(blue)(4x+2y=10

We can eliminate $8 x$ in the first equation, by $4 x$ in the second equation if we multiply it with $- 2$ (with the whole equation) to get $- 8 x$

$\rightarrow - 2 \left(4 x + 2 y = 10\right)$

Use distributive property:

color(brown)(a(b+c)=ab+ac

$\left(\mathmr{and}\right)$

color(brown)(a(b+c=x)=ab+ac=ax

$\rightarrow - 8 x - 4 y = - 20$

Add both of the equations

$\rightarrow \left(8 x - 3 y = 3\right) + \left(- 8 x - 4 y = - 20\right)$

$\rightarrow - 7 y = - 17$

color(green)(rArry=(-17)/-7=17/7

Substitute the value of $y$ to the second equation

$\rightarrow 4 x + 2 \left(\frac{17}{7} = 10\right)$

$\rightarrow 4 x + \frac{34}{7} = 10$

$\rightarrow 4 x = 10 - \frac{34}{7}$

$\rightarrow 4 x = \frac{36}{7}$

color(green)(rArrx=36/7-:4=cancel36^9/7*1/cancel4^1=9/7