How do you solve the system #14x+5y=31, 2x-3y=-29#?
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#2|k^2# then #2|k# for some #k\inZZ#"
1 Answer
Jun 12, 2018
Explanation:
Let's eliminate the y's the isolate the x and solve for x.
To do this, multiply (1) by 3, and (2) by 5.
(1)+(2) to eliminate y.
This leaves us with:
therefore
Substitute this back into one of the original equation to get:
Solve for y,
