### Video Transcript

Solve the simultaneous equations π¦ equals π₯ minus four and π₯ squared over five plus π¦ squared over three equals four, giving your answers to two decimal places.

There are several techniques that we can use to solve systems of linear equations. When the equations are of a different order β in this case, we have a linear and a quadratic β we generally tend to use the method of substitution. In this case, weβre going to substitute the equation π¦ equals π₯ minus four into the second equation. Now we might also notice that weβre told to give our answers correct to two decimal places. When weβre dealing with a quadratic equation, we tend to try and solve by factoring to find exact answers. However, since weβre told to give our answers to two decimal places, thatβs a good indication that weβre going to need to use the quadratic formula to solve any results in quadratics.

So, what does it look like when we substitute π¦ equals π₯ minus four into our second equation? Weβre going to replace π¦ squared with π₯ minus four squared. So, our second equation becomes π₯ squared over five plus π₯ minus four squared over three equals four. Now, because these fractions are making everything a little bit nasty, letβs multiply by the lowest common multiple of five and three. That will have the effect of getting rid of any fractions in this problem. π₯ squared over five times 15 is three π₯ squared, and π₯ minus four squared over three times 15 is five times π₯ minus four squared. Then, four times 15 is equal to 60.

Next, letβs distribute the parentheses π₯ minus four squared. Remember, thatβs not as simple as simply squaring the π₯ and squaring the negative four. We, in fact, need to multiply π₯ minus four by π₯ minus four. And to do so, we multiply each term in the first expression by each term in the second. π₯ times π₯ is π₯ squared; then π₯ times negative four and negative four times π₯ gives us negative four π₯ minus four π₯. Finally, negative four times negative four is 16. So, distributing the parentheses and we have π₯ squared minus four π₯ minus four π₯ plus 16, which simplifies to π₯ squared minus eight π₯ plus 16. We then replace π₯ minus four squared in our earlier equation with this new expression. And we have three π₯ squared plus five times π₯ squared minus eight π₯ plus 16 equals 60.

Letβs distribute the parentheses one more time. When we do, the left-hand side becomes three π₯ squared plus five π₯ squared minus 40π₯ plus 80. Simplifying further and our equation is eight π₯ squared minus 40π₯ plus 80 equals 60. If we then subtract 60 from both sides, we get a quadratic equation of the form we need to use the quadratic formula. Itβs eight π₯ squared minus 40π₯ plus 20 equals zero. And in fact, we could then divide everything through by four to get two π₯ squared minus 10π₯ plus five equals zero.

Letβs clear some space and use the quadratic formula to solve this quadratic equation. The quadratic formula says that the solutions to the equation ππ₯ squared plus ππ₯ plus π equals zero, if they exist, are given by π₯ equals negative π plus or minus the square root of π squared minus four ππ over two π. π is the coefficient of π₯ squared, whilst π is the coefficient of π₯, and π is the constant. So, in our quadratic equation, π is two, π is negative 10, and π is five. Substituting these values into the quadratic formula and we get π₯ equals negative negative 10 plus or minus the square root of negative 10 squared minus four times two times five over two times two. That simplifies to 10 plus or minus the square root of 60 all over four.

And so, there are two solutions to our quadratic equation, one where weβre adding 10 and the square root of 60 and one where weβre subtracting. Letβs calculate both of these values using our calculator. Correct to two decimal places, 10 plus the square root of 60 over four is 4.44 and 10 minus the square root of 60 over four is 0.56. Now it might be tempting to think weβre finished, but weβre not quite. We need to work out the corresponding values of π¦ for these two π₯-values, so we substitute each of them into the equation π¦ equals π₯ minus four. When π₯ equals 4.44 then, π¦ is 4.44 minus four, which is 0.44. Similarly, when π₯ is 0.56, π¦ is 0.56 minus four, which is negative 3.44. So, the solutions to our simultaneous equations are π₯ equals 4.44 and π¦ equals 0.44 and π₯ equals 0.56 when π¦ equals negative 3.44.

And itβs worth noting at this stage that these are pairs of solutions. The π₯-values will only work with their corresponding π¦-values and vice versa. And at this stage itβs worth noting that we can check our answer. To check, weβll substitute each pair of values into the second equation π₯ squared over five plus π¦ squared over three equals four. Weβll just demonstrate this using the first pair. When π₯ equals 4.44 and π¦ equals 0.44, the left-hand side of our equation is 4.44 squared over five plus 0.44 squared over three. And remember, if these values satisfy our equations, weβd be expecting theyβre equal to four. But we know weβve rounded our answers, so we might get a value just a little bit away from this. In fact, we get 4.007 and so on. So, we can assume that our answers are likely to be correct and weβre done.