# How do you solve the system of equations 3x + 4y = 23 and 4x - 3y = 14?

Jan 17, 2018

solve the equations (as u have only 2 variables)
else solve using determinants

#### Explanation:

multiply first equation by 3
we get $9 x + 12 y = 69$
now multiply second equation by 4
we get $16 x - 12 y = 56$
we get $25 x = 125$
hence $x = 5$
now put x = 4 in one of the equations to get y's value
hence $y = 2$
(or)
hope u find it helpful :)

Jan 17, 2018

x=5
y=2

#### Explanation:

$3 x + 4 y = 23$
4x−3y=14

$\text{First we need to get rid of either x or y}$

$3 x + 4 y = 23 , / \cdot 3$

4x−3y=14 //*(-4)

$9 x = 12 y = 69$
$16 x - 12 y = 56$

$\text{Now we add those two equations}$

$25 x = 125 / : 5$
$x = 5$

$\text{Now we need to find y, (we can use the first equation)}$
$15 + 4 y = 23 / - 15$
$4 y = 8 / : 4$
$y = 2$

Jan 17, 2018

$y = 2$ and $x = 5$.

#### Explanation:

A way of solving the system is the following:

1- Isolate $x$ for both equations:

$x = \frac{23 - 4 y}{3} \mathmr{and} x = \frac{14 + 3 y}{4}$

2- Set those two equations as equal, since they are both equivalents to $x$:

$\frac{23 - 4 y}{3} = \frac{14 + 3 y}{4}$

3- Solver this equation for $y$ by turning everything into $4$ fractions and separating fractions with y from independent fractions. $y$ should be equal to $2$

4- Now that you know that $y = 2$, plug that back in an equation of $x$, for example:

$x = \frac{23 - 4 y}{3}$

5- Once done that, you should get a value for $x$, which is $5$. Those two values are your final answer $y = 2 \mathmr{and} x = 5$

Hope this was helpful and good luck with algebra!