Simultaneous Equations

There are times when we may encounter an equation where we have two terms with unknown values. In this situation, you could be forgiven for concluding that there is no way to find the value for both. Fortunately, there is a method we can use to solve these problems. They are called simultaneous equations.

Equations that have two unknowns can seem to have an infinite number of solutions. For example in:

3 x space plus space y space equals space 11

x could equal 1, and y could equal 8

or

x could equal 2, and y could equal 5

or

x could equal 3, and y could equal 2

And that’s not accounting for decimals!

So how do we find the values of the terms in the equation above? The approach is called simultaneous equations because we use a pair of equations to find the values. Here are the equations:

 3 x space plus space y space equals space 11

2 x space plus space y space equals space 8

To use simultaneous equations, the first step is to get a common coefficient in the equations. A coefficient is a number multiplying a value. At the moment, we have a coefficient of 3 in the first equation and 2 in the second. We can get a common coefficient by multiplying these together like so:

left parenthesis cross times 2 right parenthesis space space 6 x space plus space 2 y space equals space 22

left parenthesis cross times 3 right parenthesis space space 6 x space plus space 3 y space equals space 24

And now we subtract one equation from the other:

space space space space 6 x space plus space 2 y space equals space 22
bottom enclose negative space 6 x space plus space 3 y space equals space 24 end enclose
space space space space space space space space space space space minus 1 y space equals space minus 2

Or:

y space equals space 2

Now we can input the value of 2 for y into the simplest equation from above:

2 x space plus space 2 space equals space 8

And we can rearrange the equation to isolate x by subtracting 2 from both sides of the equation then dividing both sides by 2 and  :

x space equals space fraction numerator left parenthesis 8 space minus space 2 right parenthesis over denominator 2 end fraction

And then doing the calculation:

x space equals space 6 over 2

x space equals space 3

We can use both values in our original equation to check it makes sense:

left parenthesis space 3 space cross times space 3 space right parenthesis space plus space 2 space equals 11

left parenthesis space 2 space cross times space 3 space right parenthesis space plus space 2 space equals space 8

Job done!

We can use or simultaneous equation solving skills on real-life problems of critical importance. We want to revive an old recipe for grandmother’s famous cake, but the recipes we’ve found have missing elements and some odd units of measurement. Here’s what we know:

7.5 space h a n d f u l s space f l o u r space plus space 3 space s c o o p s space s u g a r space equals space 384 space g

5 space h a n d f u l s space f l o u r space plus space 1 space s c o o p space s u g a r space equals space 253 space g

Don’t get distracted by the use of scoops and handfuls. We just need to treat them the same way we did the xs and ys earlier.

First, let’s get our common coefficient:

left parenthesis cross times 5 right parenthesis space space 37.5 space h a n d f u l s space f l o u r space plus space 15 space s c o o p s space s u g a r space equals space 1 comma 920 space g

left parenthesis cross times 7.5 right parenthesis space space 37.5 space h a n d f u l s space f l o u r space plus space 7.5 space s c o o p s space s u g a r space equals space 1 comma 897.5 space g

Now we can subtract one equation from the other:

space space space space 37.5 space h a n d f u l s space f l o u r space plus space 15 space s c o o p s space s u g a r space equals space 1 comma 920 space g
bottom enclose negative space 37.5 space h a n d f u l s space f l o u r space plus space 7.5 space s c o o p s space s u g a r equals space 1 comma 897.5 space g end enclose
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 7.5 space s c o o p s space s u g a r space equals space 22.5 space g

Now we can rearrange this to find out the weight of a scoop of sugar:

1 space s c o o p space s u g a r space equals space fraction numerator 22.5 space g over denominator 7.5 end fraction

1 space s c o o p space s u g a r space equals space 3 space g

And now we can substitute this value for sugar scoops into the simplest of our simultaneous equations:

5 space h a n d f u l s space f l o u r space plus space 3 space g space equals space 253 space g

And then rearrange the equation to isolate a handful of flour:

a space h a n d f u l space o f space f l o u r space equals space fraction numerator open parentheses 253 space g space minus space 3 space g close parentheses over denominator 5 end fraction

And calculate the value:

a space h a n d f u l space o f space f l o u r space equals space fraction numerator 250 space g over denominator 5 end fraction

 a space h a n d f u l space o f space f l o u r space equals space 50 space g

Finally, to check, we can use the values in the original equations:

left parenthesis space 7.5 space cross times space 50 space right parenthesis space plus space left parenthesis space 3 space cross times space 3 space right parenthesis space equals space 384 space g

left parenthesis space 5 space cross times space 50 space right parenthesis space plus space 3 space equals space 253 space g

We have the recipe. Let’s hope the cake was worth the effort!

To help reinforce what we have just learned, on the next page there are some practice questions. Answers and explanations can be revealed by clicking on show answer.

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