Video: Solving Simultaneous Equations Where One Is a Circle and the Other Is Linear

Solve the simultaneous equations 𝑦 = π‘₯ βˆ’ 2, (π‘₯ βˆ’ 2)Β² + (𝑦 βˆ’ 3)Β² = 9.

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Video Transcript

Solve the simultaneous equations 𝑦 equals π‘₯ minus two, π‘₯ minus two squared plus 𝑦 minus three squared is equal to nine.

The first thing to notice is that this is a linear-quadratic system of equations. The first equation 𝑦 equals π‘₯ minus two is a linear equation, as the highest power of π‘₯ and 𝑦 that appears is one. And the second equation π‘₯ minus two squared plus 𝑦 minus three squared equals nine is a quadratic equation because once we’ve distributed the parentheses, we will have both π‘₯ squared and 𝑦 squared terms.

We’re going to use the method of substitution to answer this problem. Now, our first equation is 𝑦 equals π‘₯ minus two. And we notice that the expression π‘₯ minus two appears in the second equation. So what we can do is replace π‘₯ minus two with 𝑦 in our second equation. Doing so gives 𝑦 squared plus 𝑦 minus three squared equals nine. And so we have an equation in 𝑦 only. It’s a quadratic equation which we can solve.

We begin by distributing the parentheses in 𝑦 minus three squared. And we remember that 𝑦 minus three all squared means 𝑦 minus three multiplied by 𝑦 minus three. So when we distribute the parentheses, we have four terms. And these four terms then simplify to 𝑦 squared minus six 𝑦 plus nine. Collecting like terms on the left-hand side, we have the quadratic equation two 𝑦 squared minus six 𝑦 plus nine equals nine.

Now, we notice that as we have positive nine on each side, these two terms will directly cancel one another out, which is equivalent to subtracting nine from each side of the equation. This leaves us with the simplified equation two 𝑦 squared minus six 𝑦 equals zero, which we can solve by factoring. The highest common factor of two 𝑦 squared and negative six 𝑦 is two 𝑦. And to make two 𝑦 squared, we have to multiply two 𝑦 by 𝑦. And to make negative six 𝑦, we have to multiply two 𝑦 by negative three. So our quadratic in its factored form is two 𝑦 multiplied by 𝑦 minus three is equal to zero.

To solve, we take each factor in turn, set it equal to zero, and then solve the resulting linear equation. The first equation is two 𝑦 equals zero, which we can solve by dividing each side by two to find that 𝑦 is equal to zero. The second equation is 𝑦 minus three equals zero. We solve by adding three to each side giving 𝑦 equals three. So we find that our solution to these simultaneous equations includes two 𝑦-values, 𝑦 equals zero and 𝑦 equals three.

We now need to find the corresponding π‘₯-values. And to do this, we substitute each 𝑦-value into the linear equation. It must be the linear equation into which we substitute because it’s possible that, on the quadratic curve, there is more than one point where 𝑦 equals zero or 𝑦 equals three. But on the linear graph, the straight line, there will only be one point where 𝑦 equals zero and one point where 𝑦 equals three. So we must substitute into the linear equation.

When 𝑦 equals zero, we have zero equals π‘₯ minus two. And adding two to each side, we find that π‘₯ is equal to two. When 𝑦 equals three, we have that three is equal to π‘₯ minus two. And adding two to each side, we find that π‘₯ is equal to five. So the solution to this pair of simultaneous equations is two pairs of π‘₯𝑦-values. π‘₯ equals two and 𝑦 equals zero and π‘₯ equals five and 𝑦 equals three.

It’s important to understand that these solutions come in pairs. And we can’t mix and match the π‘₯- and 𝑦-values. For example, π‘₯ equals two and 𝑦 equals three is not a valid solution to this pair of simultaneous equations, which you can confirm by substituting the values into the linear equation or, indeed, the quadratic.

Remember, we used the method of substitution to answer this question by replacing π‘₯ minus two with 𝑦. It would also have been possible to replace 𝑦 in the second set of parentheses with π‘₯ minus two to give an equation in π‘₯ only. In this case, we’d have found our solutions for π‘₯ first and then substitute it back into the linear equation to find our solutions for 𝑦. Our final answer is that π‘₯ equals two and 𝑦 equals zero or π‘₯ equals five and 𝑦 equals three.

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