### Video Transcript

Use substitution to solve the simultaneous equations one-third π₯ plus two-thirds equals π¦ and six π₯ plus three-fifths π¦ equals 64 over five.

To solve this pair of simultaneous equations means we need to determine the values of the two variables π₯ and π¦ that satisfy both equations. Weβre told that we need to approach this problem using the substitution method, which means that we need to find an expression for one variable in terms of the other and then substitute this expression into the other equation to give an equation in one variable only.

Now, considering the equations weβve been given, we notice that π¦ is in fact the subject of the first equation. So weβve been given an expression for π¦ in terms of the other variable. π¦ is equal to one-third π₯ plus two-thirds. We can therefore take this expression for π¦ and substitute it into the second equation in place of π¦. Letβs see what that looks like.

So we take the second equation, and where we had π¦ before, we now replace that with one-third π₯ plus two-thirds. So the equation becomes six π₯ plus three-fifths multiplied by one-third π₯ plus two-thirds is equal to 64 over five. And weβve now created an equation in one variable only. Our equation is in terms of π₯.

We can now solve this equation to determine the value of π₯. The first step is going to be to distribute the parentheses. In each case, the three in the numerator of the fraction weβre multiplying by will cancel with the three in the denominator of the other fraction. So weβre left with six π₯ plus one-fifth π₯ plus two-fifths is equal to 64 over five. We now notice that three of the terms involve fractions with a denominator of five. So it will make life easier if we turn this six π₯ term also into a fraction with a denominator of five. Six is of course 30 divided by five. So we can rewrite six π₯ as 30 over five π₯. And now every term in our equation has the same denominator of five.

By multiplying the entire equation through by five then, we eliminate all of the fractions. And we obtain 30π₯ plus π₯ plus two equals 64. Next, we can group the like terms on the left-hand side. 30π₯ plus π₯ is 31π₯. And then we can subtract two from each side of the equation to give 31π₯ is equal to 62. The final step is to divide each side of the equation by the coefficient of π₯, which is 31. So we have π₯ is equal to 62 over 31. And 62 over 31 is equal to two.

Weβve therefore found the value of π₯. All that remains is to find the value of the other variable π¦. We can do this by substituting the value weβve just found for π₯ into either of the two equations. As equation one gives an explicit expression for π¦ in terms of π₯, itβs perhaps easier to use this equation. Substituting π₯ equals two then gives π¦ is equal to one-third multiplied by two plus two-thirds. Thatβs two-thirds plus two-thirds, which is equal to four-thirds. So we found the solution to this pair of simultaneous equations. π₯ is equal to two, and π¦ is equal to four-thirds.

Itβs sensible to check our answer though. And we can do this by substituting the values of both π₯ and π¦ into the second equation. When we substitute these values of π₯ and π¦ into the expression on the left-hand side, we get six multiplied by two plus three-fifths multiplied by four-thirds. Six multiplied by two is of course 12. And cross canceling a factor of three from the numerator and denominator of the second product, we have that three-fifths multiplied by four-thirds is equal to four-fifths. 12 is equal to 60 divided by five. So we can rewrite this sum as 60 over five plus four over five, which is 64 over five. This is the same as the value on the right-hand side of equation two. So this confirms that our solution is correct.

Using the substitution method then, we found that the solution to the given pair of simultaneous equations is π₯ equals two and π¦ equals four-thirds.