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

multiply first equation by 3
we get `9x + 12y = 69`
now multiply second equation by 4
we get `16x - 12y = 56`
add both the obtained equations
we get `25x = 125`
hence `x = 5`
now put x = 4 in one of the equations to get y's value
hence `y = 2`
(or)
https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&cad=rja&uact=8&ved=0ahUKEwi1n4fpqd_YAhVHGZQKHU29BsoQFggzMAM&url=http%3A%2F%2Fwww.analyzemath.com%2FTutorial-System-Equations%2Fcramers_rule.html&usg=AOvVaw0WTZHPNNd0K1ISsn2SljUh
hope u find it helpful :)

`3x+4y=23`
`4x−3y=14`

`"First we need to get rid of either x or y"`

`3x+4y=23, //*3`

`4x−3y=14 //*(-4)`

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

`"Now we add those two equations"`

`25x=125 //:5`
`x=5`

`"Now we need to find y, (we can use the first equation)"`
`15+4y=23 //-15`
`4y=8 //:4`
`y=2`

A way of solving the system is the following:

1- Isolate `x` for both equations:

`x=(23-4y)/3 and x=(14+3y)/4`

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

`(23-4y)/3 = (14+3y)/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=(23-4y)/3`

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

Hope this was helpful and good luck with algebra!