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Question Video: Solving Simultaneous Equations by Elimination Mathematics • 8th Grade

Use the elimination method to solve the simultaneous equations (2/3 π‘₯) + (3/4 𝑦) = 17/24, (3/4 π‘₯) + (5/3 𝑦) = 3/2.

05:40

Video Transcript

Use the elimination method to solve the simultaneous equations two-thirds π‘₯ plus three-quarters 𝑦 equals 17 over 24, four-thirds π‘₯ plus five-thirds 𝑦 equals three over two.

Solving this pair of simultaneous equations means we need to find the values of the two variables π‘₯ and 𝑦 that satisfy both equations. We’re told that we need to use the elimination method, which means we need to find some way of eliminating or removing one of the variables from these equations so that we create a new equation in one variable only. The way we do this is by making the coefficients of either π‘₯ or 𝑦 the same in both equations. In fact, they don’t have to be exactly the same. They can be the same size or magnitude, but one can be positive and one can be negative. We’ll see why this helps in a moment.

For now, we observe that the coefficients of π‘₯ in the two equations are two-thirds and four-thirds and the coefficients of 𝑦 are three-quarters and five-thirds. If we were to multiply the entirety of equation one by the constant two, then the coefficient of π‘₯ would be two-thirds multiplied by two, which is four-thirds. And that’s the same as the coefficient of π‘₯ in the second equation. We have to multiply every term in the equation by two though, which has the effect of increasing the numerator of each fraction by a factor of two. So we have four-thirds of π‘₯ plus six over four or six-quarters of 𝑦 is equal to 34 over 24. We can label this as equation three and then rewrite equation two below it.

What we now see is that the coefficient of π‘₯ is exactly the same in the two equations. This means that if we subtract one equation from the other, then the π‘₯-terms will be eliminated because if we have any number of π‘₯ and then we subtract that same number of π‘₯, we’re left with no π‘₯’s. So, let’s subtract equation two from equation three. When we do that, the π‘₯-terms cancel as we’ve already said. We have six over four 𝑦 minus five over three 𝑦, which we could write as six over four minus five over three all multiplied by 𝑦. And on the right-hand side, we have 34 over 24 minus three over two. We can simplify some of the fractions. Six over four is equivalent to three over two, and 34 over 24 is equivalent to 17 over 12.

We then need to find common denominators in order to be able to subtract each pair of fractions. On the left-hand side, we’ll use the common denominator of six, and three over two is equivalent to nine over six, while five over three is equivalent to 10 over six. So we have nine-sixths minus ten-sixths multiplied by 𝑦. On the right-hand side, we’ll use the common denominator 12, and three over two is equivalent to 18 over 12. So we have negative one-sixth 𝑦 is equal to negative one twelfth. And by multiplying or dividing both sides by negative one, we can cancel out the negatives, leaving one-sixth of 𝑦 is equal to one twelfth. To solve for 𝑦, we multiplied both sides of the equation by six, giving 𝑦 is equal to six over 12, which is equivalent to one-half.

So, we’ve found the value of one of these two variables, and now we need to find the value of the other. To do this, we substitute the value of 𝑦 we’ve just found into any of the three equations, either equation one, two, or three. Let’s use equation one. This gives two-thirds of π‘₯ plus three-quarters multiplied by one-half is equal to 17 over 24. Three-quarters multiplied by one-half is three-eighths, so we have two-thirds of π‘₯ plus three-eighths equals 17 over 24. We can then write all of the terms in this equation with a common denominator of 24.

By multiplying both the numerator and denominator of the first fraction by eight, we see that two-thirds is equivalent to 16 over 24. And by multiplying the numerator and denominator of the second fraction by three, we see that three-eighths is equivalent to nine over 24. Multiplying the entire equation through by 24 then will eliminate all of the denominators and give the simpler equation 16π‘₯ plus nine is equal to 17. To solve this equation, we can subtract nine from each side, giving 16π‘₯ is equal to eight. We can then divide both sides of the equation by 16, giving π‘₯ equals eight over 16, which is of course equal to one-half. So, we found the solution to this pair of simultaneous equations. π‘₯ is equal to one-half, and so is 𝑦.

We should always check our answers where possible. And we can do this by substituting the values we found for π‘₯ and 𝑦 into the second equation. Substituting both π‘₯ and 𝑦 equal a half into the left-hand side of this equation gives four-thirds multiplied by a half plus five-thirds multiplied by a half, which is four-sixths plus five-sixths. This is equal to nine-sixths, and this fraction does indeed simplify to three over two, the value on the right-hand side of equation two, which confirms that our answer is correct. Using the elimination method then, we found that the solution to the given pair of simultaneous equations is π‘₯ equals a half and 𝑦 equals a half.

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