How to solve the following simultaneous equations? Equation 1: 3x+y=10 Equation 2: 2x-y=5

There are 2 easiest and quickest ways to approach this question:

1). By elimination:

When using the elimination method, you want to make sure that you choose the correct variable to eliminate to ensure only one variable (either x or y) is remained in the equation. Start by checking whether the coefficient of the variable are the same. In this case, you will notice that the coefficient of x in the two equations are not the same but the coefficient of y in the equations are both 1, which means we can choose to eliminate y. To do this, we need to check whether we are adding or subtract the equations; we can see that only one of the y in the equations is negative, suggesting we can add the equations together to eliminate y. This will result in 5x=15. Then, we can solve 5x=15, by think what times 5 equals 15, which is 3. Therefore, x=3. Many students will stop here and think that they have completed the question, however they have only solved x but not y. Now we need to find y by substituting x=3 into either equation 1 or 2. In this case, I have chosen to substitute x=3 into Equation 1 -> 33+y=10 -> 9+y=10 -> y=1. We have now found x=3 and y=1, however there is one more step we can do to ensure we have obtained the correct answer – substitute both x=3 and y=1 into the Equation 2 to double check -> LHS=23-1=6-1=5=RHS

2). By substitution:

I would only recommend to use this method if one of the equation is already in the form of x=… or y=…, otherwise you will need to rearrange the equation to make one of the variable the subject and students often lose mark during this step before they actually start solving the equations. However, I will still show you how this can be done. Firstly, you need to decide, which equation to rearrange and which variable to make the subject. Ideally you would want to choose the equations that will end up with no or less negatives and the variable to have a coefficient of 1. Therefore, I have chosen to rearrange Equation 2 and making y the subject -> y=2x-5. We can then use this to replace the y in Equation 1 -> 3x+(2x-5)=10 --(expand and simplify)-> 5x-5=10 -> 5x=15 -> x=3. Then, again we need to substitute this into equation 1 or 2 to obtain y -> 3*3+y=10 -> 9+y=10 -> y=1. Therefore, x=3 and y=1.

Glossary:

Coefficient - the number placed before and multiplying x or y

LHS - LEFT HAND SIDE of the equation

RHS - RIGHT HAND SIDE of the equation