Video: Pack 2 β€’ Paper 1 β€’ Question 23

Pack 2 β€’ Paper 1 β€’ Question 23

07:45

Video Transcript

Solve the system of equations algebraically. π‘₯ squared minus 𝑦 squared equals seven and two 𝑦 plus π‘₯ equals 10.

Let’s start by labelling the equations number one and number two. Now, we can take equation number two and make π‘₯ the subject by subtracting two 𝑦 from both sides of the equation. And this gives us that π‘₯ is equal to 10 minus two 𝑦. And we’ll call this equation three. So now, we have an equation for π‘₯ in terms of 𝑦 and we can substitute this into equation number one. This gives us 10 minus two 𝑦 all squared minus 𝑦 squared is equal to seven.

Now, we have a term of 10 minus two 𝑦 all squared, which we can expand using the FOIL method. First, we can write 10 minus two 𝑦 all squared as 10 minus two 𝑦 times 10 minus two 𝑦. Now, we can use the FOIL method. So we start by multiplying the first terms in each bracket. So that’s 10 times 10 which is equal to 100. Next, we multiply the outside terms β€” so that 10 timesed by minus two 𝑦 giving us minus 20𝑦. Next, we multiply the inside term. So that’s minus two 𝑦 times 10 which gives us minus 20𝑦. And then, finally, we multiply the last term in each bracket. So that’s minus two 𝑦 times minus two 𝑦. And since a negative times a negative gives a positive, this gives us four 𝑦 squared.

Now, we notice that we have minus 20𝑦 minus 20𝑦, which we could write as minus 40𝑦. And we get 100 minus 40𝑦 plus four 𝑦 squared. And we can substitute this back into the equation for 10 minus two 𝑦 all squared. Next, we can combine the terms with the same powers of 𝑦. So we have four 𝑦 squared minus 𝑦 squared, giving us three 𝑦 squared. Then, we have minus 40𝑦, which we don’t have anything to combine with. So we’re just left with minus 40𝑦.

Finally, we have 100 and seven. But the seven is on the other side of the equation. So we need to subtract seven from both sides of the equation, which gives us minus seven on the left of the equation. And we are left with 100 minus seven, which gives us 93. And then since we subtracted seven from the left, we also subtract seven from the right of the equation. So this means that this is all equal to zero. And we are left with a quadratic in 𝑦, which we can solve to find 𝑦.

In order to factorize this equation, we can start by drawing our brackets. And we know that in each bracket, there will be some lot of 𝑦 plus some constant since we have a 𝑦 squared term. So let’s write the 𝑦s in each bracket. Next, we notice that we have three 𝑦 squared. In order to make this three 𝑦 squared term, we will multiply the two 𝑦 terms from the brackets. So therefore, there must be a coefficient in front of one of the 𝑦 terms in order to give us the three. And since three is a prime number, this number has to be three. So we know that one of the 𝑦 terms in one of the brackets is three 𝑦. And when multiplying this by the 𝑦 in the other bracket, this will give us the three 𝑦 squared.

Next, we can work out the signs of the constants. Since we have a positive constant in the equation and a negative 𝑦 term in the equation and since the constant term comes from multiplying the last term in each bracket, this means that the constants must have the same symbol β€” so either be positive or both be negative. Now since we have a minus 𝑦 term, this is how we know that both of them are negative or else we would have no negatives in our equation.

In order to work out what the two constants are, let’s look at the possible factor pairs of 93. We either have one and 93 or three and 31. We need to find a factor pairing such that when we multiply one of the factors by three and add it to the other one, we end up with 40. So if we tried multiplying this one by three and adding it to 93, this gives us 96. So therefore, it can’t be this pairing. Next, let’s try one plus three times 93. This gives us 280. So therefore, this can’t be a factor pairing either. So we know that the first one is not right. Next, we can try three timesed by three plus 31 and this gives us 40. Therefore, we found our factor pairing.

Now, since we know that we need three times three plus 31, therefore, the three has to go in the opposite bracket to the three 𝑦. Since then when we expand it, we will end up multiplying the three 𝑦 by the three to give us the nine, which we then add to the 31 multiplied by the 𝑦 and then this is what gives us the negative 40𝑦.

So now that we have factorized our 𝑦 equation, we can find the solutions of 𝑦. And since we have two things multiplied by one another equal to zero, this means that either one of them has to be equal to zero. So three 𝑦 minus 31 equals zero or 𝑦 minus three equals zero. Now, we can solve the equation on the left for 𝑦 by adding 31 to either side, giving us that three 𝑦 is equal to 31. And then, we can divide both sides by three and this gives us that 𝑦 is equal to 31 over three. Now, we can solve the equation on the right by simply adding three to both sides and we get another solution of 𝑦 is equal to three.

So now, we have two possible solutions of 𝑦 and we can substitute them into equation three, which is π‘₯ is equal to 10 minus two 𝑦, in order to find the corresponding π‘₯-values. If we substitute in 𝑦 is equal to 31 over three, we get π‘₯ is equal to 10 minus two timesed by 31 over three. Then, two timesed by 31 over three is equal to 62 over three. And then, we know that we can write 10 as 10 timesed by three over three since three over three is equal to one and 10 timesed by three over three is equal to 30 over three. And now, since 30 over three and 62 over three have a common denominator, we can combine the fractions and we get 30 minus 62 all over three. And this simplifies to give us an π‘₯-value of minus 32 over three.

Next, we can find the corresponding π‘₯-value when 𝑦 is equal to three. Substituting 𝑦 into the equation, we get that π‘₯ is equal to 10 minus two times three. Then, two times three is equal to six. So we get 10 minus six, which is the same as four.

So now, we have found a solution to the system of equations. And the solution is 𝑦 is equal to 31 over three and π‘₯ is equal to minus 32 over three or 𝑦 is equal to three and π‘₯ is equal to four.

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