We treat this problem as two simultaneous equations, using our knowledge that when two graphs intersect, they simultaneously have the same solution. The second equation (quadratic equation) is already in the form where it is equal to y, meaning that we can substitute (x2 - 8x +7) into the first linear equation, where we obtain 2x + x2 - 8x + 7 = 7.(Equally, we could have rearranged the linear equation into the form y = 7 - 2x and substituted this into the quadratic equation to obtain the equivalent equation x2 - 8x +7 = 7 - 2x).Now we can rearrange this equation to make it into an easily solvable quadratic equation equal to zero:2x + x2 - 8x + 7 = 7--> x2 - 6x = 0Now, we can see that this can be simply solved using factorisation, so we obtain:x(x - 6) = 0, which therefore implies that there are two solutions to x: x = 0, x = 6, as when we substitute either of these solutions into the quadratic equation we obtain 0, as required.Using this information, we are able to substitute these values into either original equation to obtain two corresponding solutions for y. For example, using the linear equation:2(0) + y = 7--> y = 7, so one of the points of intersection is (0,7).Using the same method for x = 6, we get the second point to be (6,-5).(Note: we would obtain the same solutions if we had substituted the values of x into the second equation).