### Video Transcript

Solve π₯ equals π¦ minus five over two, π₯ squared minus π¦ equals 10. You must show your working. Do not use trial and improvement.

Weβve been given a pair of simultaneous equations in π₯ and π¦, which we need to solve. Weβre told that we must not use trial and improvement, which means we need to use a formal algebraic method. Letβs look at the equations weβve been given more closely.

The first equation is a linear equation because it just has π₯ and π¦ to the power of one. The second equation is a quadratic equation because it includes an π₯ squared term. In order to solve a pair of equations where one is a quadratic, we need to use the method of substitution.

Now, we could substitute π₯ equals π¦ minus five over two from equation one into equation two. But this will be quite complicated because it will involve fractions. What weβll do instead is rearrange equation one to give π¦ in terms of π₯ because this wonβt involve any fractions and then we can substitute this into equation two instead.

The first step in rearranging equation one is to multiply both sides by two as this will eliminate the denominator of two on the right-hand side. This gives two π₯ equals π¦ minus five. The next step is to add five to each side of the equation, giving two π₯ plus five equals π¦. So weβve rearranged to give π¦ in terms of π₯.

Weβre now going to substitute this expression for π¦ in terms of π₯ into equation two. Making this substitution gives π₯ squared minus two π₯ plus five equals 10. Itβs important that we have the brackets around two π₯ plus five because weβre subtracting both of these terms from π₯ squared.

To remove the bracket, we have to multiply both terms inside it by negative one which just has the effect of changing their signs. So we have π₯ squared minus two π₯ minus five equals 10. We can then subtract 10 from each side of this equation to give π₯ squared minus two π₯ minus 15 equal zero.

Now, we have a quadratic equation in π₯ only. So we need to solve it, which could be done in various ways. We could use the quadratic formula or we could complete the square. But letβs see if this quadratic will factorise first.

As the coefficient of π₯ squared is just one, weβll have π₯ at the front of each bracket. Weβre then looking for two numbers that add to the coefficient of π₯ β thatβs negative two β and multiply to the constant term β thatβs negative 15. The factors of 15 are one and 15 or three and five. And as we need these numbers to multiply to a negative number, they need to have different signs: so one needs to be positive and one needs to be negative.

If we make the three positive, but the five negative and then add three and negative five together, these two numbers do have a sum of negative two. So these are the numbers that weβre looking for, this means that our quadratic factorises as π₯ plus three multiplied by π₯ minus five. You can check this by expanding the brackets, perhaps using the FOIL method if you wish.

To solve for π₯, we then set each bracket in turn equal to zero. So we have π₯ plus three equals zero or π₯ minus five equals zero. Each equation can be solved in a relatively straightforward way. For the first equation, we subtract three from each side, giving π₯ equals negative three. For the second equation, we add five to each side, giving π₯ equals five.

So we have two values for π₯. And this will usually be the case if one of the equations that weβre solving is a quadratic. Now, we need to find the values of π¦ that pair with these values of π₯.

We can substitute each value of π₯ back into any of our equations. But by far, the more straightforward one to use will be the equation π¦ equals two π₯ plus five. When π₯ is equal to negative three, we have two multiplied by negative three plus five. Two multiplied by negative three is negative six. So we have negative six plus five which is equal to negative one. Our first pair of values then is π₯ equals negative three, π¦ equals negative one.

When π₯ is equal to five, we have π¦ equals two multiplied by five plus five which is equal to 10 plus five which is 15. Our second pair of values then is π₯ equals five, π¦ equals 15. Remember that these values do belong in pairs. We canβt mix and match them. π₯ equals five, π¦ equals negative one, for example, would not be a valid solution to this pair of equations.

We can check each pair of values by substituting back into equation two if we wish. For example, if we substitute the first pair of values back into equation two, we get negative three squared minus negative one. Negative three squared is positive nine and subtracting negative one is the same as adding one. So we have nine plus one. This is equal to 10. And as this is the right-hand side of equation two, this shows that our first pair of values are correct. You can check the second pair of values in exactly the same way.