How do i solve the simultaneous equations 7x+y=1 and 2x^2-y=3?

First, label the equations 1 and 2:1: 7x+y=1 and 2: 2x^2-y=3Rearrange equation 2 in terms of yy=2x^2-3Now substitue the rearranged version of equation 2 into 1 to give7x+2x^2-3=1.Bring all the values to the left-hand side and place the terms in indice order2x^2+7x-4=0 and call this equation 3.Now, factorise the equation. When there is a coefficient in front of the x^2, multiply your constant (the -4) by the coefficient to give you the value you need to have in your factorisation, giving you -8. In this case, you need two numbers which, when added together, give 7 and when multiplied together give -8:+=7 and x= -8The values that satisfy both of these equations are 8 and -1.Go back to euqation 3 and rewrite the 7x term as 8x-x to give2x^2+8x-x-4=0.You can now factorise this equation as2x(x+4)-(x+4)=0.Group the (x+4) terms together to give(2x-1)(x+4)=0.Since this equation has to equal 0, either x=1/2 or x=-4.Substitute x=1/2 into equation 2 to give7(1/2)+y=17/2+y=2/2y=-5/2.Now substitute x=-4 into equation 2 to give7(-4)+y=1-28+y=1y=29.You have your two solutions which you should write as coordinates (1/2, -5/2) and (-4, 29).You can check these solutions by substituting them into equation 1 to see if they work.