Solve Linear Equations -

An equation is a numerical statement that has an ‘=’ sign.

Example:

$x+5=16$

One of the numbers has been written as a letter. We solve an equation when we calculate the value of the letter.

Since we know that $11+5=16$ we can say that $x=11$ . The number $11$ is the solution to the equation above.

Equations come in many forms.

Linear equations are equations that we solve by adding, subtracting, multiplying or dividing only.

Solve a ‘one step’ equation

A one-step equation can usually be solved by number recall/logical thinking. Multi-step need the ‘completing and balancing’ method. This applet illustrates the logical thinking and the completing and balancing methods.

applet link

Solve a ‘two step’ problem

An equation such as $3x+5=26$ is often referred to as a two step problem. The value $x$ has been multiplied by 3, and then 5 has been added. Two operations. That requires two reverse operations to figure out $x$ . Here are three techniques for figuring out the value $x$ :

Guess and check. This is a good approach if the unknown is a simple whole number.
Work backwards. This is a good approach for two step equations, but not for more complicated equations.
Completing and balancing. This is what algebra is named for. Its when we “do the same to both sides” to keep the balance. This method works for all kinds of complicated equations.

applet link

Video: Mathantics Scroll to ‘Algebra Basics Part 1’

Printable worksheet: corbettmaths equations

Online interactive worksheet: transum maths equations

Solve Equations with letters on both sides

It can be helpful to follow some predetermined steps to solve these.

Expand any brackets, simplify each side if necessary;
Add or subtract the $x$ term from one side so that there is an $x$ term on one side only;
Continue as before for two step equations.

There could be a better approach than these three steps – it really depends on the equation you’re solving. Its neat to figure out your own way to solve things.

applet link

Printable worksheet from Corbettmaths, letters both sides

Online interactive exercise on solving equations letters both sides by Transum maths

When there is one fraction

$Thinking about solving linear equations with fractions$

A fraction is an alternate way to express division.

$\frac{9}{4}=9 \div 4 = 2.25$

The inverse operation of division is multiplication:

$\frac{9}{4}\times 4 =9\div 4 \times 4 = 9$

Multiplying by the denominator ‘clears’ the denominator. For example, solve

$\frac{58-x}{5}=2x+5$

In this case, the denominator is 5. Remember to multiply both sides by 5.

$\begin{align*} 5\times \Big(\frac{58-x}{5}\Big)&=5\times \Big(2x+5\Big)\\[10pt] \frac{5(58-x)}{5}&=10x+25\\[10pt]\frac{\cancel{5}(58-x)}{\cancel{5}}&=10x+25\\[10pt] 58-x&=10x+25\\[10pt]58&=11x+25\\[10pt]33&=11x\\[10pt]x&=3\end{align*}$

Alternatively, you might notice that when there is one fraction the denominator multiplies all other terms in the equation:

This saves a whole bunch of lines!

When there is more than one fraction

One method to handle an equation with more than one fraction is to:

…multiply the whole equation by the least common multiple of all denominators

First notice that we can clear a fraction by multiplying by the denominator, or by any multiple of the denominator:

$3\times \Big(\frac{2}{3}\Big)=\cancel{3}\Big(\frac{2}{\cancel{3}}\Big)=2$

Also, because $30\div 3 = 10$ :

$30\times \Big(\frac{2}{3}\Big)=^{10}\cancel{30}\Big(\frac{2}{\cancel{3}}\Big)=20$

When we need to clear several fractions at once, we multiply by a common multiple of all denominators.

$\frac{x+9}{4}-\frac{2x+7}{6}=\frac{x+7}{3}$

The least common multiple of $4, 6$ and $3$ is $12$ .

$\begin{align*}12\Big(\frac{x+9}{4}-\frac{2x+7}{6}\Big)&=12\Big(\frac{x+7}{3}\Big)\\[10pt]12\Big(\frac{x+9}{4}\Big)-12\Big(\frac{2x+7}{6}\Big)&=12\Big(\frac{x+7}{3}\Big)\\[10pt]^3\cancel{12}\Big(\frac{x+9}{\cancel{4}}\Big)-^2\cancel{12}\Big(\frac{2x+7}{\cancel{6}}\Big)&=^4\cancel{12}\Big(\frac{x+7}{\cancel{3}}\Big)\\[10pt] 3(x+9)-2(2x+7)&=4(x+7) \\[10pt] 3x+27-4x-14&=4x+28\\[10 pt] -x+13&=4x+28\\[10pt]13&=5x+28\\[10pt]-15&=5x\\[10pt] x&=-3 \end{align*}$

$x=-3$ substituted to the original line gives:

$\frac{6}{4}-\frac{1}{6}=\frac{4}{3}$

which we can verify is correct.

Printable textbook equations with fractions from corbettmaths

Online interactive exercise on solving equations with fractions from transum math

Video on the special case of (fraction) = (fraction) by Brian McLogan: Cross Multiply

Check out these online/interactive/printable learning resources.

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