Equations

Lesson

This topic builds on what you learnt in One Step, Two Step. The aim of the game hasn't changed. We are still trying to solve for unknowns and to do so, we need to make the unknown the subject of the equation (which means it's on one side on the equals sign by itself).

You can think of solving an equation as reversing a series of operations to find the value of the unknown. In three step equations, these operations may involve a combination of addition, subtraction, multiplication and division to make the variable the subject of the equation (in other words get it to one side of the equation by itself).

Consider the equation $\frac{3x-1}{4}=2$3`x`−14=2. In this equation, what are the operations to reverse?

Well, $x$`x` has been:

1. Multiplied by 3

2. Then 1 has been subtracted

3. Then the result has been divided by 4

We would need to reverse each of these three operations to solve for the value of $x$`x`.

Remember that whatever operation we do to one side we must do to the other.

Solve: $2\left(4x+4\right)=56$2(4`x`+4)=56

Consider the equation $\frac{3x}{5}-1=2$3`x`5−1=2.

Solve for the value of $x$

`x`that satisfies the equation.To verify the solution graphically, which two straight lines would need to be graphed?

$y=\frac{3x}{5}$

`y`=3`x`5A$y=\frac{3x}{5}+1$

`y`=3`x`5+1B$y=2$

`y`=2C$y=\frac{3x}{5}-1$

`y`=3`x`5−1D$y=\frac{3x}{5}$

`y`=3`x`5A$y=\frac{3x}{5}+1$

`y`=3`x`5+1B$y=2$

`y`=2C$y=\frac{3x}{5}-1$

`y`=3`x`5−1DGraph the lines $y=\frac{3x}{5}-1$

`y`=3`x`5−1 and $y=2$`y`=2 on the same plane.Loading Graph...Hence find the value of $x$

`x`that satisfies the two equations $y=\frac{3x}{5}-1$`y`=3`x`5−1 and $y=2$`y`=2 simultaneously.

Consider the equation $2\left(x-1\right)-3=7$2(`x`−1)−3=7.

Solve for the value of $x$

`x`that satisfies the equation.To verify the solution graphically, which two straight lines would need to be graphed?

$y=2\left(x-1\right)+3$

`y`=2(`x`−1)+3A$y=7$

`y`=7B$y=2\left(x-1\right)-3$

`y`=2(`x`−1)−3C$y=2\left(x-1\right)+4$

`y`=2(`x`−1)+4D$y=2\left(x-1\right)+3$

`y`=2(`x`−1)+3A$y=7$

`y`=7B$y=2\left(x-1\right)-3$

`y`=2(`x`−1)−3C$y=2\left(x-1\right)+4$

`y`=2(`x`−1)+4DGraph the lines $y=2\left(x-1\right)-3$

`y`=2(`x`−1)−3 and $y=7$`y`=7 on the same plane.Loading Graph...Hence find the value of $x$

`x`that satisfies the two equations $y=2\left(x-1\right)-3$`y`=2(`x`−1)−3 and $y=7$`y`=7 simultaneously.