3.02 Review: One-step equations

Keeping equations balanced

We want to keep equations balanced so that the two sides of the equals sign remain equivalent. If we don't we could change what the equation means. Think of a balanced set of scales. The scale remains level when the weights on both side of the scales are even. The same thing happens with equations.  

If we add a weight to one side and not to the other, then the scales will no longer be balanced. 

Exploration

This applet represents the equation $x=5$x=5. What equivalent equations can you make by doing the same thing to both sides? See if you can come up with $4$4 different equations. 

Worked examples

Question 1

Beginning with the equation $x=8$x=8, write the new equation produced by adding $2$2 to both sides.

Think: We will start with $x=8$x=8 and then to keep the equation balanced, we must do the same thing to both sides.

Do:

$x$x	$=$=	$8$8	Given equation
$x+2$x+2	$=$=	$8+2$8+2	Adding $2$2 to both sides
$x+2$x+2	$=$=	$10$10	Simplify - perform the addition

 

Question 2

Beginning with the equation $x=-99$x=−99, write the new equation produced by dividing both sides by $11$11.

Think: We will start with $x=-99$x=−99 and then to keep the equation balanced, we must do the same thing to both sides.

Do:

$x$x	$=$=	$-99$−99	Given equation
$x\div11$x÷​11	$=$=	$\left(-99\right)\div11$(−99)÷​11	Dividing both sides by $11$11
$\frac{x}{11}$x11​	$=$=	$-9$−9	Simplify - write as a fraction and perform the division

 

Question 3

Beginning with the equation $33x=99$33x=99, write the new equation produced by dividing both sides by $11$11.

Think: We will start with $33x=99$33x=99 and then to keep the equation balanced, we must do the same thing to both sides.

Do:

$33x$33x	$=$=	$99$99	Given equation
$33x\div11$33x÷​11	$=$=	$99\div11$99÷​11	Dividing both sides by $11$11
$3x$3x	$=$=	$9$9	Simplify - perform the division

 

 

 

Practice questions

Question 4

Question 5

Solving a one-step equation

We can often solve for an unknown value by setting up an equation and then solving for the unknown value, often represented with a variable. To solve for the unknown variable, we will use opposite operations to get it by itself.

Worked example

Question 6

Suppose a person holding a dog steps on a scale. The number that shows on the scale is $75$75 kg. The person weighs $70$70 kg. First set up an equation for this scenario using $x$x as the weight of the dog, then solve the equation.

Think: We know that $\text{Person }+\text{Dog }=75$Person +Dog =75, that $\text{Person }=70$Person =70, and that $\text{Dog }=x$Dog =x. Using this we can set up an equation.

Do: 

$70+x$70+x	$=$=	$75$75	Filling in what we know
$x+70$x+70	$=$=	$75$75	We often write the variable first
$x+70-70$x+70−70	$=$=	$75-70$75−70	Subtract $70$70 from both sides to get $x$x by itself
$x$x	$=$=	$5$5	Simplify

 

The dog weighs $5$5 kg.

Reflect: What other strategies could you use to find the weight of the dog? Does this answer make sense?

 

 

Practice questions

Question 7

Question 8

Question 9