solve this question 2x+3y=8, x^2-xy+yz=3 by factorization

Question

Solve by factorization 2x+3y=8 X^2-xy+yz=3

Answer 1

For 3 variables x, y, z, you need 3 equations to get a unique solution. As it stands we can only find two variable in terms of the third.

Multiply x²-xy+yz=3 by 3:

3x²-3xy+3yz=9 and substitute 3y=8-2x:

3x²-8x+2x²+8z-2xz=9,

5x²-2x(4+z)+8z-9=0,

x=(2(4+z)±√(4(4+z)²-160z+180))/10,

x=(4+z±√(16+8z+z²-40z+45))/5,

x=(4+z±√(z²-32z+61))/5. We don't know z so we can't find x or y. The quadratic in z has real solutions only when z²-32z+61≥0.

One set of solutions is when z=2, so x=7/5 or 1, making y=26/15 or 2.