Can you help me to solve the simultaneous equation plzzzz???

Question

2y-x=1
XY=x2=26

Answer 1

The second thing you entered isn't an equation. What you probably mean is
xy - x^2 = 26.

If that is the second equation, what you'd do is solve this by substitution. You'd solve the first equation for "x", so it would say
2y - 1 = x

Then substitute "2y - 1" everywhere there's an "x" in the second equation. You'd get
(2y - 1)y - (2y - 1)^2 = 26

Distribute everything out. You get
2y^2 - y - (4y^2 - 4y + 1) = 26
2y^2 - y - 4y^2 + 4y - 1 = 26
-2y^2 + 3y - 1 = 26
0 = 2y^2 - 3y + 27

You would most likely use the quadratic formula to find possible values of "y". Then for each value of "y", you'd substitute back to find corresponding values of "x".