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

3 Answers
Jan 17, 2018

Answer:

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

Explanation:

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 :)

Jan 17, 2018

Answer:

x=5
y=2

Explanation:

#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#

Answer:

#y=2# and #x=5#.

Explanation:

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!