Solving 1-variable equations

TL;DR

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Desmos is a perfectly acceptable method for solving many linear equations, but it may be overkill on easier questions. You should, therefore, get really quick at solving equations with one variable as these question types make up a decent portion of the early questions on the test.

tip

Abstraction is a powerful way to skip steps confidently and become faster at solving equations.

Keep the variable positive

Given this equation:

$$2x-5=8x-11$$

If you move the x to the left, you’ll wind up dealing with a lot of negatives, which can easily lead to mistakes:

$$2x-5=8x-11$$ $$-6x-5=-11$$ $$-6x=-6$$ $$x=1$$

This is better:

$$2x-5=8x-11$$ $$-5=6x-11$$ $$6=6x$$ $$1=x$$

Skip steps by switching terms

One great trick for increasing your speed and accuracy simultaneously is to switch terms. Consider this equation:

$$10-x=7$$

You could simply move the 10 to the right, but this does require handling more negative numbers, which in turn leads to mistakes.

Instead, you should simply switch the positions of the x and the 7, like so:

$$10-x=7$$ $$10-7=x$$ $$3=x$$

In general, given an equation in the form:

$$a-b=c$$

We can switch the b and the c to get:

$$a-c=b$$

We can also do this with division. For example, given this equation:

$$5=\frac{7}{x}$$

We can simply switch the 5 and x to get our final solution:

$$x=\frac{7}{5}$$

In general, given an equation in the form:

$$a=\frac{b}{c}$$

We can switch the a and the c to get:

$$c=\frac{b}{a}$$

Avoid expanding brackets (if possible)

Many students reflexively expand brackets as a first step — but this is often a tactical mistake. Instead, you should try to rearrange the equation with breaking it up into smaller pieces. For example, give this equation:

$$10=2(x-3)$$

It would be a tactical mistake to multiply the 2 into the brackets, as this leads to extra steps, and extra steps mean more chances of messing up. Instead, the best move is first divide both sides by two:

$$\frac{10}{2}=\frac{2(x-3)}{2}$$

And solve from there:

$$5=x-3$$ $$8=x$$

Here’s another example where expanding brackets is a tactical mistake:

$$5(x+1)-3(x+1)=8$$

Instead of expanding, we can note that the (x+1) factor is common to both terms, so we simply subtract 3 from 5 to get 2:

$$2(x+1)=8$$

Now again we DON’T expand the brackets. Instead we divide both sides by 2:

$$\frac{2(x+1)}{2}=\frac{8}{2}$$

And solve

$$x+1=4$$ $$x=3$$