Solve the following simultaneous equations: x^2-y^2=9, x - y = 1
We first notice we have a linear and non linear equation. The simplest method to solve is to substitute the linear equation into the non linear equation and then solve for one of the unknowns. Then substitute that value once known back into on the of the equations (into the linear one is simpler) and then solve again for the second unknown.
x - y = 1 ==> x = 1 + y
x^2-y^2=9 ==> (1 + y) ^2 - y ^ 2 = 9 ==> 2y = 8 ==> y = 4
x = 1 + y ==> x = 1 + 4 ==> x = 5
x = 5
y = 4
SH
Related Maths GCSE answers
All answers ▸The Curve C has the equation 2x^2-11+13. The point Q lies on C such that the gradient of the normal to C at Q is -1/9. Find the x-co-ordinate of Q
Solve algebraically the simultaneous equations: x^2 + 3x + 10 = y and; 4 - 2x = y. Give answers as co-ordinates.
X is inversely proportional to Y. When X=4, Y=9. If Y=6, what is the value of X?
Solve this quadratic equation 3x(squared)-7=101