Video Transcript
Find all of the solutions to the simultaneous equations 𝑦 plus nine 𝑥 plus seven equals zero and 𝑥 squared plus 𝑦 squared minus seven 𝑥𝑦 equals four. Give values to two decimal places.
There are a number of techniques we can use to solve a system of linear equations. However, we have one linear equation: 𝑦 plus nine 𝑥 plus seven equals zero. That’s an equation whose graph is simply a straight line. And we have a quadratic equation. This involves 𝑥 squared and 𝑦 squared terms. And when we solve these systems of equations, the most sensible method to choose is substitution.
We’re going to form an expression for either 𝑥 or 𝑦 with our first equation and then substitute it into our second. So let’s inspect this first equation: 𝑦 plus nine 𝑥 plus seven equals zero. We want to find a way to make either 𝑦 or 𝑥 the subject. It follows that since the coefficient of 𝑦 is one, in other words there is simply one 𝑦 in the equation, we should make 𝑦 the subject. To do so, we’re going to subtract nine 𝑥 and seven from both sides of the equation. And we see that this equation can be alternatively written as 𝑦 equals negative nine 𝑥 minus seven.
Now, if we define this equation to be equation one and our quadratic to be equation two, we’re going to substitute equation one into equation two. And to do so, we simply replace every instance of 𝑦 with the expression negative nine 𝑥 minus seven and then manipulate using the order of operations. And so when we do, equation two becomes 𝑥 squared plus negative nine 𝑥 minus seven squared minus seven 𝑥 times negative nine 𝑥 minus seven equals four. And we might notice that when we distribute any parentheses, we’re going to have a quadratic equation in terms of 𝑥.
There is a hint as to how we’re going to solve that quadratic equation in the question. It tells us to give values to two decimal places. That is an indication to us that the quadratic equation will not be easily solved by factoring. And so we’re probably going to end up using the quadratic formula.
Before we do though, let’s distribute the parentheses and set this equation equal to zero. Let’s multiply out the expression negative nine 𝑥 minus seven squared. That’s negative nine 𝑥 minus seven multiplied by itself. Then, we multiply the first term in each expression to get 81𝑥 squared. Next, negative nine 𝑥 times negative seven and negative seven times negative nine 𝑥 gives us two lots of 63𝑥. Finally, negative seven times negative seven is 49. And we’ve distributed our parentheses. And this simplifies to 81𝑥 squared plus 126𝑥 plus 49.
Then, our earlier equation becomes 𝑥 squared plus this expression plus the result of distributing negative seven 𝑥 across negative nine 𝑥 minus seven. That’s 63𝑥 squared plus 49𝑥. And of course this is equal to four. Let’s collect like terms on the left-hand side first. We have 145𝑥 squared and we have 175𝑥. Remember, we want to set this expression equal to zero. So we’re going to subtract four as well. And so this equation is 145𝑥 plus 175𝑥 plus 45 equals zero.
And whilst not entirely necessary, it can be sensible to simplify even further wherever possible. Here, each term contains a common factor of five. So we divide everything through by five. And the equation that we need to solve is 29𝑥 squared plus 35𝑥 plus nine. And remember, we said that we can solve this equation using the quadratic formula. This formula tells us that the solutions to the equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero are 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 over two 𝑎. Now, 𝑎 is the coefficient of 𝑥 squared, so it’s 29; 𝑏 is the coefficient of 𝑥, it’s 35; and 𝑐 is the constant term, it’s nine.
Substituting these values into our quadratic formula, and we get 𝑥 equals negative 35 plus or minus the square root of 35 squared minus four times 29 times nine over two times 29. And that simplifies to negative 35 plus or minus root 181 over 58. So there are two possible values of 𝑥: one which is negative 35 plus the square root of 181 over 58 and the other is negative 35 minus the square root of 181 over 58. Correct to two decimal places, they are negative 0.37 and negative 0.84, respectively.
Now, in fact, we’re not quite finished. We need to work out the corresponding 𝑦-values for these 𝑥-values. To do so, we substitute them back into equation one. And for accuracy, we’ll use the exact values. So we’ll work out 𝑦 equals negative nine times negative 35 plus root 181 over 58 minus seven. Correct to two decimal places, that gives us a 𝑦-value of negative 3.66. Then, we’ll repeat that process for our second value of 𝑥, once again using the exact value. And that gives us, correct to two decimal places, 0.52.
And so we have all of the solutions to the simultaneous equations. Using set notation, that’s the set containing the ordered pairs negative 0.37, negative 3.66 and negative 0.84, 0.52.
