Solve the simultaneous equations algebraically
x2+y2=25 and y-3x=13Step 1 is to label the equations (1) and (2). We shall label the first equation, the quadratic, as equation (1) and the second equation, the linear, as equation (2). When solving a simultaneous equations with 2 unknowns we are looking to remove one of the variables in place of the other variable. To do this, we can rearrange (2) such that we are left with y=13+3x. If we square both sides of this equation we are left with y2=(13+3x)2=(13+3x)(13+3x). Using the FOIL method (First, outside, inside, last) we can multiply out these two brackets to get an equivalent value, such that y2=9x2+78x+169. We can now substitute this value of y2 into (1) to give us x2+(9x2+78x+169)=25 therefore 10x2+78x+144=0.
Now that we have one equation with only one unknown we can factorise it to find the value of x. To make this easier, we can simplify the quadratic as all coefficients are a factor of 2. Therefore we can divide both sides of the equation by 2. An easy way to see if we can simplify an equation is to check what numbers go into the smallest coefficient and then check them on the larger coefficients. We are now left with 5x2+39x+72=0. There are four methods to factorise a quadratic equation, factoring, completing the square, using the quadratic formula and graphically. As the coefficients are relatively friendly we can use factoring. If we notice that 5 and 1 are the only numbers that will go into 5 we know that the 'x' coefficients will be 5 and 1. The two remaining integers in each bracket are 24 and 3, as 24 times 3 gives us 72. We can see that 24 + (3x5) gives us 39 as well so we know that the factorised form for our equation is (5x+24)(x+3)=0. From this we know that 5 x + 24 = 0 therefore x = -24/5 or x + 3 = 0 therefore x = -3. We can substitute these values for x into a rearranged form of equation (2), y=13+3x such that when x = -24/5, y = -7/3 or when x = -3, y = 4
