solve the simultaneous equation; x^2+y^2=10 2x+y=5

When dealing with a linear and quadratic simultaneous equation, it is best to use subsitution to approach it.I take the linear equation and put it in terms of one variable. In this case, to avoid fractions i will put everything in terms of y.2x+y-2x=5-2x (i take away 2x from each side so the equation does not change but it is in terms of y)y=5-2x (as 2x-2x=0)Now we can use this value for y in the quadratic equation so that we can factorise it.x²+(5-2x)²=10 (if you know how to quickly expand a squared number with variables it would be an advantage but i will to it both ways)PERFECT SQUARE METHOD(x+a)²=x²+2ax+a²a=5 x=-2x in this case(-2x)²+2(-2x)(5)+(5)² (Multiplying it out you get )4x²-20x+25MULTIPLYING IT OUT(5-2x)(5-2x) ..... as any number squared is the number multiplied by itself5(5-2x)-2x(5-2x) ...to mutliply it fully out you need to mutlipy each term in one bracket by the terms in another bracket.25-10x-10x+4x² ..... putting like terms together.25-20x+4x²4x²-20x+25 = (5-2x)²Back to the equation, sub 4x²-20x+25 for (5-2x)²x²+4x²-20x+25=10 .... minus 10 from each side to make an equation to factorise and put like terms together5x²-20x+15=0 ..... since 5 is a common mutliple in each term then divide each term by 5x²-4x+3=0 factorise(x-3)(x-1)=0x-3=0 x-1=0x=3 x=1now that we got our two values for x, we will get two values for y by subbing the x vaules into the orginal equationsy=5-2(3) y=5-2(1)y=5-6 y=5-2y=-1 y=3now our answer is x=3 y=-1, x=1 y=3