Question #f621d

1 Answer
Azimet
Jan 20, 2018

If #p+q=4#, then #y# can take any value. Otherwise, #y=0#.

Explanation:

It's a bit weird for the problem to ask to "solve for #y#" because if we tried to do so, we'd get "#y=0#, provided #p+q!=4#" without a way to express #y# in terms of #p,q#...anyway,

we have #py+qy=-4y+8y#. Let's factor out #y# from both sides to make things simpler:

#(p+q)y=4y#. Then we can subtract #4y# from both sides, and factor out #y# again from the left:

#(p+q-4)y=0#.

This is true either when #p+q=4#, in which case #y# can be anything, or when #y=0#.

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