Question Video: Solving Simultaneous Using Elimination, Where One of the Equations Needs to be Multiplied | Nagwa Question Video: Solving Simultaneous Using Elimination, Where One of the Equations Needs to be Multiplied | Nagwa

Question Video: Solving Simultaneous Using Elimination, Where One of the Equations Needs to be Multiplied Mathematics • Third Year of Preparatory School

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Using elimination, solve simultaneous equations 5𝑥 − 4𝑦 = 21, 4𝑥 + 12𝑦 = 32.

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Video Transcript

Using elimination, solve simultaneous equations five 𝑥 minus four 𝑦 equals 21, four 𝑥 plus 12𝑦 equals 32.

So we’re asked to solve this system of equations using the elimination method, which means we’re looking to eliminate either the 𝑥- or 𝑦-variable by adding or subtracting our two equations. However, if we were to try this as the equations currently are, we’d find that in both cases we still have 𝑥- and 𝑦-variables in the equation we are left with. If we were to add, we’d have the equation nine 𝑥 plus eight 𝑦 equals 53. And if we were to subtract, we’d have the equation 𝑥 minus 16𝑦 equals negative 11. So this hasn’t actually helped.

So why hasn’t it worked? Well, in order to use the method of elimination, we’re looking for the coefficients of one of the variables to be the same in both equations, or at least to have the same magnitude such as positive and negative three. But in this problem, this isn’t the case. We have a coefficient of five for 𝑥 in the first equation and four in the second. And we have a coefficient of negative four for 𝑦 in the first equation and 12 in the second. So simply adding or subtracting these equations as they currently are doesn’t eliminate either variable.

We’re told, though, that we need to use the method of elimination. So what should we do? Well, what we’re going to do is manipulate these equations slightly so that we do have the same coefficient, or at least the same magnitude of coefficient for one of the variables. To achieve this, we use the equality property of multiplication, which says that if we multiply both sides of an equation by the same value, then the equality is still true. We’re looking for a value that we can multiply one equation by which will then give the same coefficient, or same-size coefficient, for one of the variables in both equations.

Looking at the 𝑦-variables in our two equations, we see that four is a factor of 12. So if we were to multiply equation 1 by three, then we would have negative 12𝑦. And so the magnitude of the coefficients of 𝑦 would be the same in both equations. Let’s try that then. Let’s multiply equation 1 by three. So we multiply everything in equation 1 by three, giving 15𝑥 minus 12𝑦 is equal to 63. The coefficients of 𝑦 in our two equations are now the same size but with different signs, which means we can eliminate the 𝑦- variables by adding our two equations together.

When we do, we have 15𝑥 plus four 𝑥, which gives 19𝑥; negative 12𝑦 plus 12𝑦, which gives zero; and on the right-hand side 63 plus 32, which is 95. So we’ve eliminated the 𝑦-variable from our equations, giving a single equation in 𝑥, which we can solve by dividing both sides by 19. Doing so gives the solution for 𝑥. 𝑥 is equal to five.

We can then find the value of 𝑦 by substituting 𝑥 equals five into any of the three equations, either of the original two or the equation we created when we multiplied equation 1 by three. I’m going to use equation 2 as all of the coefficients are positive in this equation. Doing so gives four times five, which is 20, plus 12𝑦 equals 32. We can then subtract 20 from each side and divide by 12 to give 𝑦 equals one. So we have our solution: 𝑥 equals five and 𝑦 equals one. But of course, we should check it. I’m going to choose to check by substituting the values into equation 1. That’s five 𝑥 minus four 𝑦 equals 21. Substituting 𝑥 equals five and 𝑦 equals one gives five times five minus four times one. That’s 25 minus four, which is indeed equal to 21. So this confirms that our solution is correct.

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