Video: Pack 2 β€’ Paper 2 β€’ Question 16

Pack 2 β€’ Paper 2 β€’ Question 16

06:29

Video Transcript

Solve the following simultaneous equations using algebra: π‘₯ is equal to two 𝑦 minus seven over four, π‘₯ squared minus one over 16 is equal to two 𝑦 squared.

First, we’ll number the two equations as equation one and equation two so that we have some shorthand to use in referring to them. Now, note that equation one is the simpler equation. It’s a linear equation, meaning that the powers of π‘₯ and 𝑦 are both one, whereas equation two is a quadratic β€” the powers of π‘₯ and 𝑦 are two.

Equation one gives us an expression for π‘₯ in terms of 𝑦. So we can substitute this expression for π‘₯ into the second equation. And this will give an equation in terms of 𝑦 only, making this substitution gives two 𝑦 minus seven over four all squared minus one over 16 is equal to two 𝑦 squared. And therefore, there’re no π‘₯ terms in this equation.

We want to solve this quadratic equation for 𝑦. And the first step is we need to expand the bracket on the left-hand side. Remember when we’re squaring a bracket, this actually means that we’re multiplying the bracket by itself. We can use the FOIL method to help us with this expansion.

Multiplying the first terms in the brackets together, we have two 𝑦 multiplied by two 𝑦 which gives four 𝑦 squared. Multiplying the outer terms of the brackets together, we have two 𝑦 multiplied by negative seven over four. The two in the numerator of this product and the four in the denominator can be cancelled by a factor of two, leading to negative seven 𝑦 over two.

Next, we multiply the inner terms in the brackets together: negative seven over four multiplied by two 𝑦 which is the same product as we’ve just done, but in reverse. So it simplifies to the same thing, negative seven 𝑦 over two. Finally, we multiply the last term in each bracket together: negative seven over four multiplied by negative seven over four which gives positive 49 over 16.

When we add the four terms in the expansion together negative seven 𝑦 over two plus another lot of negative seven 𝑦 over two gives negative seven 𝑦. So the simplified expansion is four 𝑦 squared minus seven 𝑦 plus 49 over 16.

Returning then to our solution to the simultaneous equations, we now have four 𝑦 squared minus seven 𝑦 plus 49 over 16 minus one over 16 is equal to two 𝑦 squared. Now, 49 over 16 minus one over 16 is equal to 48 over 16. And our 16 times three is 48. This just simplifies to three. We’ll also group all the terms on the left-hand side of the equation by subtracting two 𝑦 squared from both sides. This gives two 𝑦 squared minus seven 𝑦 plus three is equal to zero.

Now, this is a quadratic equation in 𝑦 only. And to solve, we have a number of possible methods: we could factorize, we could use the quadratic formula, or we could complete the square. If a quadratic does factorize, then it’s almost always quickest to use this method. So it’s always worth a quick check first.

In this quadratic, the coefficient of 𝑦 squared and the constant term are both prime numbers. So we don’t have a lot of options that we need to try. The two 𝑦 squared must be the product of two 𝑦 and 𝑦. So this gives the first term in each bracket. The only factors of three are three and one. And as the constant term is positive, but the coefficient of 𝑦 is negative, this tells us that both values in the brackets must be negative. So we need negative one in one of the brackets and negative three in the other. And we can use trial and error to see which gives the correct factorization.

Now, I’ve put the negative one in the first bracket and the negative three in the second for now. And we need to check that when we expand these brackets, we get the correct number of 𝑦s. We already know that we’ll get two 𝑦 squared and we’ll get positive three. So we only need to check the middle two terms in the expansion. The 𝑦 terms come from multiplying two 𝑦 by negative three giving negative six 𝑦 and negative one by 𝑦 giving negative 𝑦. In total then, we have negative six 𝑦 plus negative 𝑦 giving negative seven 𝑦 which is what we wanted.

So we have the correct factorization for this quadratic and we’re nearly there with solving for 𝑦. Remember if two brackets multiplied to give zero, then at least one of the brackets must itself be equal to zero. So we have that two 𝑦 minus one is equal to zero or 𝑦 minus three is equal to zero. These are linear equations that we can solve to find 𝑦.

To solve the first equation, we add one and then divide by two, giving 𝑦 is equal to one-half. To solve the second equation, we just need to add three to each side giving 𝑦 is equal to three. So we found the values of 𝑦 that satisfy the simultaneous equations. And the final step is we need to find the values of π‘₯.

To do so, we’ll substitute into the more straightforward equation β€” equation one. When 𝑦 is equal to a half, π‘₯ is equal to two multiplied by a half minus seven over four. Two multiplied by a half is one. So we have one minus seven over four. Now, one can be expressed as four over four. So we have four over four minus seven over four which is negative three over four.

When 𝑦 is equal to the other value three, π‘₯ is equal to two multiplied by three minus seven over four which gives six minus seven over four. Six can be written as 24 over four. So we have 24 over four minus seven over four which is 17 over four.

The solution then to the simultaneous equations are π‘₯ equals negative three-quarters, 𝑦 equals a half and π‘₯ equals 17 over four, 𝑦 equals three.

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