How do you solve the following linear system 4x−3y=−5,4x±4y=−4?

1 Answer
Leland Adriano Alejandro
Feb 28, 2016

For equations 4x−3y=−5 and 4x+4y=−4, solution (−87,17)
For equations 4x−3y=−5 and 4x−4y=−4, solution (−2,−1)

Explanation:

We will use method of elimination of variables

For equations 4x−3y=−5 and 4x+4y=−4

4x+4y=−4
4x−3y=−5

Perform subtraction to obtain

7y=1

then y=17
Using 4x+4y=−4 and y=17

We have, 4x+4(17)=−4

x+17=−1

x=−87

For equations 4x−3y=−5 and 4x+4y=−4, solution (−87,17)

Again, we will use method of elimination of variables

For equations 4x−3y=−5 and 4x−4y=−4

4x−4y=−4
4x−3y=−5

Perform subtraction to obtain

y=−1

then y=−1
Using 4x−4y=−4 and y=−1

We have, 4x−4(−1)=−4

x−(−1)=−1

x+1=−1

x=−2

For equations 4x−3y=−5 and 4x−4y=−4, solution (−2,−1)
graph{(4x-3y+5)(4x+4y+4)(4x-4y+4)=0 [-3, 2, -5, 5]}

God bless....I hope the explanation is useful..

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