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1.01 Linear equations

Lesson

A linear equation is one in which each term is a constant or a variable to the power of 1. Examples of linear equations in one variable are:

$2x+3=10$2x+3=10,         $5y-2=3\left(2y+5\right)$5y2=3(2y+5),         $\frac{4z+6}{5}=10z+1$4z+65=10z+1

Later in this chapter we will look at non-linear equations, such as:

$x^2+5x-6=0$x2+5x6=0 (quadratic)   and   $2x^3-8x=0$2x38x=0 (cubic)

When we want to solve linear equations, we are trying to find an unknown value. This is achieved by rearranging the equation using inverse operations to isolate the unknown.

While rearranging an equation, it is very important that at each stage we keep the equations equivalent or balanced. This means that whatever operation we did to one side of the equation, we had to do to the other.

In some questions, we have algebraic terms on both sides of the equation. To solve these, we need to get all the algebraic terms on one side and all the numbers on the other. Then we can solve the equation as usual.

Practice question

Question 1

Solve the following equation: $2\left(2x+5\right)=3\left(x+5\right)$2(2x+5)=3(x+5)

 

Problem solving with equations

Once you are able to group pronumerals and solve equations, you can use these skills to solve problems in context.

Practice question

Question 2

Kate and Isabelle do some fundraising for their sporting team. Together they raised $\$600$$600.

If Isabelle raised $\$p$$p, and Kate raised $\$272$$272 more than Isabelle:

  1. Solve for the value of $p$p.

  2. Now calculate how much Kate raised.

Working with fractions

Here we are looking at solving equations with fractions. Once again we will follow the reversed order of operations and make the pronumeral the subject of the equation.

It may be helpful to write any fractions as improper fractions rather than mixed numbers. We can also remove fractions from the equation by multiplying both sides by the lowest common denominator.

Remember!

You can check your solution by substituting it back into the original equation.

Practice questions

Question 3

Solve the following equation:

$\frac{14k-76}{2}-2=2k$14k7622=2k

Question 4

Solve $\frac{x+2}{2}+3=\frac{x+3}{7}$x+22+3=x+37 for $x$x.

 

Outcomes

ACMMM005

solve linear equations

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