Solve the simultaneous equations, x^2+y^2=41, and y=2x-3

STEP 1) First, substitute 2x-3 into the quadratic equation, replacing y, which works out as x²+(2x-3)²=41STEP 2) Now, expand the bracket, (2x-3)², firstly by writing it in an easier way, (2x-3)(2x-3), then expand to 4x²-6x-6x+9STEP 3)Once the same powers have been grouped together for the full equation, ie all x² 's together, all x's together and all constants (normal numbers) together, you should have 5x²2-12x+9=41STEP 4)To solve this quadratic equation it needs to be equal to 0, so you need to subtract 41 from both sides, 5x²-12x-32=0STEP 5)To solve this you can use the quadratic formula, where a=5, b=-12 and c=-32, then put these numbers into the formula (I will go through the quadratic formula on the whiteboard)From using the quadratic formula your answers should be x=4 and x=-1.6STEP 6) You now need to substitute these values back into the original equation, y=2x-3, to find out the value of y isWhen x=4y=2(4)-3, = 5ANDWhen x=-1.6y= 2(-1.6)-3, = -6.2STEP 7) To check that you are right you can put your answers into the original quadratic equation, x²+y²=41, (4²)+(5²)= 16+25=41 (CORRECT)(-1.6²)+(-6.2²)= 2.56+38.44= 41 (CORRECT)