Solve algebraically the simultaneous equations: x^2 + y^2 = 25 and y - 3x = 13
Firstly we should look at the question and notice that the two possible methods are substitution or elimination. We should then notice that since one equation is a quadratic and the other linear, then the elimination method wouldn't work hence we should focus on the substitution method. We then look at the two equations and which would be easier to substitute into the other and so by observation we will notice that substituting into the quadratic equation will be easier. Now we know our method we will then calculate it.
So, as y - 3x = 13, then y = 3x + 13
Substituting y into the quadratic equation we get, x2 + (3x + 13)2 = 25
Expanding, simplifying and making it equal to zero we get, 10x2 + 78x + 144 = 0
Factorising we get, 2(5x + 24)(x + 3) = 0
So we obtain the solutions for x as, x = -24/5 and x = -3
So the final step is substituting the values for x we have obtained back into one of the original equations so that we will obtain the corresponding value for y which solves the simultaneous equation. We should choose to substitute back into the linear equation since it is a simpler substitution.
So we have, y = 3(-24/5) + 13 and y = 3(-3) + 13
So we get, y = -7/5 and y = 4 respectively
So the solutions to the simultaneous equations are x = -24/5, y = -7/5 and x = -3, y = 4
