### Video Transcript

Find all the real solutions to the following system of equations, π¦ is equal to negative two π₯ squared plus 12π₯ minus 13, π¦ is equal to negative π₯ squared over three plus two π₯ plus two.

To find all the real solutions to these two equations, weβre looking to find the values for π₯ and π¦ which satisfy our equations. And to do this, we spot that the subject of each equation is π¦. The first equation tells us that π¦ is equal to negative two π₯ squared plus 12π₯ minus 13. And the second equation tells us that π¦ is also equal to negative π₯ squared over three plus two π₯ plus two.

Remember, since we are trying to find the value of π¦ which satisfies both equations, we can say that this means that negative two π₯ squared plus 12π₯ minus 13 must be the same as, must be equal at some point to, negative π₯ squared over three plus two π₯ plus two. And now we have a quadratic equation.

We know this is a quadratic equation because its order is two. The highest power, or highest exponent, of π₯ is two. But how do we solve a quadratic equation? Well, we need to ensure that we have some expression in terms of π₯, which is then made equal to zero. There are a number of ways we can achieve this.

Weβre going to begin by multiplying everything in this equation by three to get rid of that nasty fraction. When we do, on the left-hand side, weβre left with negative six π₯ squared plus 36π₯ minus 39. And then, on the right-hand side, remember, that negative π₯ squared divided by three and then multiplied by three is simply negative π₯ squared. So, we have negative π₯ squared plus six π₯ plus six.

Weβre then going to deal with the smallest number of π₯ squared. Thatβs negative six π₯ squared. Weβre going to add that to both sides of our equation. And on the left-hand side, that leaves us with just 36π₯ minus 39. On the right-hand side, we get five π₯ squared plus six π₯ plus six. Weβre going to subtract 36π₯, leaving us with negative 39 on the left-hand side and five π₯ squared minus 30π₯ plus six on the right.

Finally, we add 39 to both sides of the equation. And weβre left with that zero we needed on the left-hand side. And on the right, we have five π₯ squared minus 30π₯ plus 45. And once our quadratic equation is equal to zero, our next step is usually to factor the quadratic expression five π₯ squared minus 30π₯ plus 45.

Itβs much easier to simplify as far as possible though. And here, we notice that each term in our equation has a factor of five. So, we divide through by five. And we can do this because our equation is equal to zero, and zero divided by five is still zero. And that leaves us with zero is equal to π₯ squared minus six π₯ plus nine. And now we want to factor the expression π₯ squared minus six π₯ plus nine.

We know this is going to factor into two brackets, since each term in this expression shows no other common factor aside from one. And we know that the number at the front of each bracket must be π₯, since when we distribute our brackets, we multiply the first term in each bracket. And π₯ multiplied by π₯ is that π₯ squared that we need.

To find the numerical term that goes in each bracket, we need to find two numbers that multiply to make nine and add to make the coefficient of π₯, thatβs negative six. So, we begin by listing the factor pairs of nine. At first, it might seem that the only factor pairs of nine are one and nine, and three and three. Remember though, a negative multiplied by a negative is a positive. So, there are two extra factor pairs for nine then, negative one and negative nine, and negative three multiplied by negative three.

So, which ones of these give us a sum of negative six? Negative three multiplied by negative three is nine. And negative three plus negative three is negative six, which is what we were after. And so, the numerical parts of our brackets is negative three. And we fully factorise our expression. π₯ squared minus six π₯ plus nine is the same as π₯ minus three multiplied by π₯ minus three. So, what next?

Well, we have two brackets for which their product is zero. So, what does that mean about each of the brackets? Well, what that means is that either bracket must be equal to zero. The only way to multiply two numbers and get a solution of zero is if at least one of those numbers is zero. Now since these brackets are identical, we can say that simply π₯ minus three must be equal to zero. And then, we solve this as normal. Weβll add three to both sides of the equation. And we see that weβve solved the equation for π₯. We have π₯ is equal to three.

Weβre not quite finished though. Remember, we said we needed to find the values of π₯ and π¦ that satisfied both equations. So, weβre going to take this value of π₯ and weβre going to substitute it into either one of the original equations. Now it doesnβt really matter which equation we choose, but Iβm going to choose negative two π₯ squared plus 12π₯ minus three purely because thereβs no fractions in it.

Substituting π₯ is equal to three into this equation, and we get π¦ is equal to negative two multiplied by three squared plus 12 multiplied by three minus 13. And remember, PEMDAS tells us to deal with the exponent before multiplying. Three squared is nine. So, π¦ is given by negative two multiplied by nine, which is negative 18, plus 12 multiplied by three, which is 36, minus 13, which is five.

π₯ is equal to three and π¦ is equal to five. And we can check these answers. To check them we substitute our value of π₯ into the other equation. This time, thatβs negative three squared over three plus two multiplied by three plus two. Thatβs negative nine over three, which is negative three, plus two multiplied by three, which is six, plus two, which, once again, gives us an answer of five. And thatβs a good indicator that what weβve done is probably correct.