There are several approaches to solving problems with simultaneous equations. Let’s begin with the standard methods that are taught in schools:

**1) Substitution**

This is generally more difficult than cancellation, but it has the advantage of working on certain types of problems for which the other method does not.

Pick one of the equations, and solve for one variable in terms of the other. Then substitute that value in the other equation and solve.

Q. If 3x + 2y = 19, and 2x + 4y = 18, what is the value of x + y?

3x + 2y = 19 Subtract 3x from both sides of the question to get

2y = 19 – 3x Divide both sides by 2 to get

y = (19 – 3x)/2 Now substitute into the other equation (2x + 4y = 18) and get

2x + 4[(19 – 3x)/2] = 18 (see – it gets a bit rough) Simplify

2x + 2(19 – 3x) = 18 Expand

2x + 38 – 6x = 18 Combine like terms

4x + 38 = 18 Subtract 38 from both sides

-4x = -20 Divide both sides by -4 and

x = 5 Now substitute 5 for x in either equation and solve for y

3(5) + 2y = 19

15 + 2y = 19

2y = 4

y = 2

So x + y = 5 + 2 = 7

Yes, that’s messy, but if you’re comfortable with the algebra, and patiently write out each step, you’ll be okay. Now let’s do the other “school technique” to solve the same question:

**2) Cancellation**

Q. If 3x + 2y = 19, and 2x + 4y = 18, what is the value of x + y?

The idea is to cancel out the x or the y terms by multiplying one of the equations so that the terms are the same. I can see that if you double the 2y term in the first equation, it will be the same as the 4y term in the second equation.

3x + 2y = 19 (multiply both sides by 2)

2x + 4y = 18

6x + 4y = 38

2x + 4y = 18

Now just subtract the second equation from the first:

(6x + 4y = 38)

-(2x + 4y = 18)

4x = 20

x = 5, and proceed as in method 1)

Many teachers say to multiply by -2 and add the two equations, but I hate negative signs because they breed careless errors (subtracting the equations uses fewer negative signs here).

As you can see, that’s easier than substitution. However, it’s actually messier if there are exponents, if some variables are multiplied and not added, or if there are more than 2 equations. However, those exceptions are rare, and the following shortcuts can usually be used:

**3) Plugging In**

As terrific a technique as plugging in is, you usually can’t use it on simultaneous equations, because you’ll back yourself into a corner (i.e. your values will work for only one of the equations). However, you can try it if you’re asked to solve for one variable in terms of another:

To plug in, substitute some simple numbers into the problem, solve it, and then substitute into the answers (you’re looking for the same value that you got when you solved the problem).

Q. If x = 3y + 1 and z = 9y^{2}, which of the following gives the value of z?

a) 3x^{2}

b) 3x

c) x + 1

d) (x + 1)^{2}

e) (x – 1)^{2}

^{ }You can plug in – say y = 2. Then you get x = 3(2) + 1 = 6 = 1 = 7, and z = 9y^{2 }= 9 X 2^{2} = 9 X 4 = 36. If you try x = 7 in each answer, only e) = 36.

**4) Backsolving**

Backsolving (substituting the answers into a variable in the question) works far more often on these problems. However, it will not work on the question shown for the first two methods, because you were asked to find x + y (backsolving only works when you’re asked to find a single variable). Let’s modify that question slightly:

Q. If 3x + 2y = 19, and 2x + 4y = 18, what is the value of x?

a) 2

b) 3

c) 4

d) 5

e) 6

When backsolving, start with the middles value (c). So substitute x = 4 into the first equation. Solve and get y = 3.5. Now try your values for x and y in the second equation: 2(4) + 4(3.5) = 18. But 8 + 14 = 22, so answer c) is incorrect. Try another answer – in d), x = 5. Solving the first equation, y = 2. Substitute those values into the second equation: 2(5) + 4(2) = 18. Yup, 10 + 8 = 18, so d) is the answer.

That takes a bit of work, but it’s very easy work.

**5) Guess and Check**

G&C only works on easy questions involving simultaneous equations.

Q. If x + y = 10, and xy = 16, then x – y could equal

a) 3

b) 4

c) 5

d) 6

e) 8

Just try some numbers that add up to 10 until you find a pair whose product is 16. 1 and 9 – no. 2 and 8 – yes.

**6) Add ’em All**

This is a nifty trick that useful when the question asks for more than one variable.

Q. If 2a + b + c = 11, a + 2b + c = 13, and a + b + 2c = 12, what is the value of a + b + c?

First, line ’em up, and then add ’em all:

2a + b + c = 11

a + 2b + c = 13

a + b + 2c = 12

4a + 4b + 4c = 36

If you simplify (divide both sides of the equation by 4), you get a + b + c = 9.

Sometimes, you need to “subtract ’em all”:

Q. If 5x + 7y = 64 and 6x + 8y = 74, what is the value of x + y?

Subtract the first equation from the second:

6x + 8y = 74

5x + 7y = 64

x + y = 10

That’s it. Unless you’re a “master algebraist,” learn how to use methods 3 – 6 (methods 3 – 5 will be useful on many other types of questions as well), and you’ll prosper on these questions.