# How do you solve the system 5a+3b=17 and 4a-5b=21?

Nov 17, 2015

To solve a system of linear equations that are both in standard form, the most efficient method to use would be the "elimination method".

#### Explanation:

To carry out elimination, you need to alter one (or both) of the equations so as to have one of the variables (and their coefficient) match the same one in another equation.

1) $5 a + 3 b = 17$
2) $4 a - 5 b = 21$

Say I want to solve for b, this means I need to eliminate a. To do this, I need to multiply eq. 1 by 4 and eq. 2 by 5 like so:

1) $4 \left[5 a + 3 b = 17\right]$
2) $5 \left[4 a - 5 b = 21\right]$

This results in:

1) $20 a + 12 b = 68$
2) $20 a - 25 b = 105$

Now, if the variable you are trying to eliminate has the same sign in both equations, you subtract the equations. If the signs are opposite, you add them. To remember this, think OASS (opposite, add, same, subtract). In this case, the signs are the same so you want to subtract the equations, resulting in:

$\left(20 a - 20 a\right) - \left(12 b - 25 b\right) = \left(68 - 105\right)$

$37 b = - 37$

$b = - 1$

Now that you have b, you can substitute that back into one of the original equations to solve for a. I'll sub it into eq. 1:

$5 a + 3 \left(- 1\right) = 17$

Solving for a:

$5 a = 20$

$a = 4$

Therefore, your solution set (or point of intersection) is (4, -1).

Hopefully, this was all clear and I hope you found it helpful! :)