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)
