Solve the simultaneous equations 'x-2y=3' and 'x^2+2y^2=27'
Equation 1: x-2y=3Equation 2: x2+2y2=27The easiest way to do this is to put one of these equations into the other, by rearranging for either x or y. Because there are no squares to deal with in equation 1, we will rearrange this for x: so x-2y=3 becomes x=3+2y, which we'll call equation 3.Equation 3: x=3+2yWe know that this is now going to be put into equation 2 as x2, so we first should first square all of equation 3. This can be done by putting them in brackets and multiplying out: x2=(3+2y)(3+2y)=4y2+12y+9. We have now got rid of x, and equation 3 is now in a form that can be easily put into equation 2: 4y2+12y+9+2y2=27. In order to solve for y, this equation must be rearranged so it equals 0, to factorise it. Rearranging the equation gives us 6y2+12y-18=0. A factor of 6 can be taken out to simplify the equation further, which gives us y2+2y-3=0. This is now in a form that we can factorise, with the numbers in the brackets needing to: add to 2 (which is the number before the y in the main equation); and multiply to -3 (which is the number at the end of the equation). This then gives us: (y+3)(y-1)=0. So we have our answers to y: -3 and 1! We can now use these answers for y to find x by putting them in equation 3: x=3+2y.So when y=-3, x=3+2(-3)=-3 and when y=1; x=3+2(1)=5. Final answer: x=5; y=1 and x=-3; y=-3
