### Video Transcript

Find the set of points of intersection of the graphs 𝑥 plus 𝑦 equals eight and 𝑥 squared plus 𝑦 squared is equal to 50.

In order to solve this pair of simultaneous equations, we firstly need to consider the linear equation 𝑥 plus 𝑦 equals eight. Rearranging this equation by subtracting 𝑥 from both sides of the equation gives us 𝑦 equals eight minus 𝑥. Our next step is to substitute this new equation into equation two. 𝑥 squared plus 𝑦 squared is equal to 50. This gives us 𝑥 squared plus eight minus 𝑥 squared equals 50.

We can expand or simplify eight minus 𝑥 all squared using the FOIL method. Multiplying the first terms in the parentheses or brackets gives us 64. Multiplying the outside terms gives us negative eight 𝑥. Multiplying the inside terms also gives us negative eight 𝑥. And finally, multiplying the last terms gives us positive 𝑥 squared. This means that eight minus 𝑥 all squared is equal to 64 minus 16𝑥 plus 𝑥 squared.

Putting this back into our equation gives us 𝑥 squared plus 64 minus 16𝑥 plus 𝑥 squared equals 50. Grouping the like terms gives us two 𝑥 squared minus 16𝑥 plus 64 equals 50. And finally subtracting 50 from both sides of this equation gives us two 𝑥 squared minus 16𝑥 plus 14 equals zero.

Before we try and factorize this quadratic equation, we can divide by two, as all of the coefficients are even. Dividing by two gives us 𝑥 squared minus eight 𝑥 plus seven equals zero. Factorizing this quadratic gives us 𝑥 minus seven multiplied by 𝑥 minus one. Therefore, our two values of 𝑥 are 𝑥 equals seven or 𝑥 equals one.

In order to find the corresponding values of 𝑦, we need to substitute these values of 𝑥 back into equation one. Eight minus seven is equal to one. And eight minus one is equal to seven. Therefore, 𝑦 could be equal to one or 𝑦 could be equal to seven. This means that the two points of intersection of the graphs 𝑥 plus 𝑦 equals eight and 𝑥 squared plus 𝑦 squared equals 50 are seven, one and one, seven. We could check these answers by substituting both of the coordinates back into the equations.