ALGEBRA

Solving Equations

Solving Equations

Inverses

MathsWhen you get an algebraic equation each letter stands for a number. In order to solve an equation you need to figure out the value of the missing letter.

One way to solve an equation is to use inverses. Basically this means that you undo the equation.

Here are a few inverses:

n?

+

x

an

For example: solve the equation 3b +7 = 13

First of all you subtract 7 from each side (the inverse of plus 7):

3b + 7 – 7 = 13 – 7

3b = 6

Now, you divide each side by 3 (the inverse of multiply 3):

3b/3 = 6/3

b = 6/3 = 2

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Brackets

If you get an equation with brackets and need to solve it remember to multiple out the brackets first.

For example, solves 2(h + 4) = 16

First of all, multiply out the brackets:

2h + 8 = 16

Now, subtract 8 from both sides:

2h = 8

Then finally, divide each side by 2:

h = 4

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Solving inequalities

When you solve an equality you don’t just get one answer but a whole range of answers.

For example the equation 4x + 2 = 10 is not the same as 4x + 2 ? 10.

For the first equation there is only one answer. For the inequality equation the answer could be one of a lot of answers.

For instance: solve 4x + 2 ? 10.

First of all, make x the subject as you would with a normal equation:

4x + 2 ? 10

4x ? 8

x ? 2

This means that the equation is satisfied when x is larger than 2.

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Double inequalities

You might be asked to solve a double inequality. This is where you get three parts to the equation.

Just remember that whatever you do to one part you must do to all the others too.

For example: solve -8 ? 3x + 2 ? 11

Minus 2 to from each part:

-6 ? 3x ? 9

Now, divide all parts by 3:

-2 ? x ? 3

So, the inequality is satisfied when x is between -2 and 3, including 3.

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Unknowns on both sides

You might also get a question in which you have unknowns on both sides.

For example: 5c + 6 = c + 8. Find the value of c.

First of all, minus 6 from both sides (the smaller number by the larger):

5c = c + 2

Now, subtract c from both sides:

4c = 2

Now, divide both sides by 4 and you’ll get your answer:

c = 2/4 = 1/2

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Trial and improvement

mathsIf you’re not given values for an algebraic equation you might need to try and estimate what they are. This is known as trial and improvement.

For example, answer the following equation to one decimal place: x2 + 6x = 25.

So, you need to figure out what number could possibly replace x.

Your first try can be a guess. So, try 4:

42 + (6 x 4) = 16 + 24 = 40

So, this answer is too big. This means you should try with a smaller number, for example x = 2:

22 + (6 x 2) = 4 + 12 = 16

So answer is too small. This means you need to try a number in-between like x = 3:

32 + (6 x 3) = 9 + 18 = 27

This answer is a little bit too large. So the third try could be: x = 2.5:

2.52 + (6 x 2.5) = 6.25 + 15 = 21.25

This is a bit too small. Fourth try: x = 2.8:

2.82 + (6 x 2.8) = 7.84 + 16.8 = 24.64

This is the closest yet. Fifth try: x = 2.85:

2.852 + (6 x 2.85) = 8.1225 + 17.1 = 25.2225

This means x is greater than 2.8 but smaller than 2.85 so, to one decimal place, the answer is 2.8.

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Equations and fractions (Higher tier)

For the higher tier you’ll be expected to solve equations which contain fractions.

There are two main methods you can use:

  • – multiply out the fraction first
  • – multiply out the fraction at the end

For example, solve 6 – x/4 = 3

If you multiply out the fraction first then you multiply everything by 4:

24 – x = 12

Then add the x:

24 = 12 + x

Lastly subtract the 12:

12 = x

Or, you can multiply out the fraction at the end:

6 – x/4 = 3

6 = 3 + x/4

3 = x/4

12 = x

Always remember to do the same to both sides.

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Simultaneous equations (Higher Tier)

mathsYou’ll also need to know how to work out simultaneous equations. this is where you have two equations with unknowns. the name of these equations comes from the fact that you should end up solving both equations at the same time.

For example:

Solve these equations:

1) 2xy + y = 10

2) x + y = 4

Take the simpler equation, which is number 2 and substitute it into number 1.

In order to do this you need to rearrange it so that either y or x is the subject. As there are more y’s in (1) you should make y the subject. This makes equation 3:

x + y = 4

3) y = 4 – x

Now you can substitute (3) into (1):

2x(4 -x) + (4 – x) = 10

8x – 2x2 + 4 – x = 10

8x – 2x2 + 4 – x – 10 = 0

7x – 2x2 – 6 = 0 or -2x2 + 7x – 6 = 0

You can swap the signs to get rid of the minus at the front:

2x2 – 7x + 6 = 0

Now you can put the equation into brackets:

(2x – 3) (x – 2) = 0

From this you can now say that:

2x – 3 = 0 or x – 2 = 0

Therefore, x = 1.5 or x = 2

Dont forget to work out the values of y aswell! From equation 3:

y = 4 – x

y = 2.5 or y = 2

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Linear equations

It’s also possible to solve simultaneous equations using a graph.

If you use this example:

1) y = 2x – 2

2) 3y = 2x + 6

First of all pick two easy numbers to plot, for example where x = 0 and where y = 0.

If x = 0 in (1) then y = -2. On your graph you can then plot (0, -2).

When y = 0 in (1) then x = 1 so you then plot (1, 0) and connect these two together to give your first line.

You now need to calculate points for the other line:

If x = 0 in (2) then y = 2. On your graph you can then plot (0, 2).

When y = 0 in (2) then x = -3 so you then plot (-3, 0) and connect these two together to give your second line.

Where the lines cross are your solutions to x and y.

Only use this method if asked: it’s much faster to use algebra otherwise.

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