Values that satisfy equations II

We've learnt how to determine whether or not a value satisfies an equation. To determine this, we need to check whether the left hand side (LHS) of the equation is equal to the right hand side (RHS) of the equation. However, sometimes an equation may not just have one answer- sometimes there may be an infinite number of solutions and sometimes there may be no solutions.

For example, what happens if we had an equation $10=-4$10=−4? We know this is not a solution. In fact, this is an example of an equation that has no solutions.

What happens if we have an equation $x=x$x=x? If we substituted values into this equation, we could say $1=1$1=1, $2=2$2=2, $3=3$3=3 and so on. This is an example of an equation that has an infinite number of solutions.

If we get a solution to an equation such as $x=5$x=5, this means that $5$5 is the only value that satisfies the equation. It is example of an equation with one solution.

Examples

Question 1

Consider the equation $16+4x=6x+6\times4$16+4x=6x+6×4.

a) Solve for $x$x.

Think: We need to collect the like terms, the simplify and solve for $x$x.

Do:

$16+4x$16+4x	$=$=	$6x+6\times4$6x+6×4
$16$16	$=$=	$2x+24$2x+24
$-8$−8	$=$=	$2x$2x
$-4$−4	$=$=	$x$x
$x$x	$=$=	$-4$−4

 

b)  How many solutions does $16+4x=6x+6\times4$16+4x=6x+6×4 have?

Think: How many solutions did we find in part a?

Do: There is one solution to this equation.

 

Question 2

How many solutions does $\frac{10+x}{10}=\frac{x+10}{10}$10+x10​=x+1010​ have?

Think: How can we transfer the given equation into a simpler form ($x=a$x=a,$a=a$a=a or $a=b$a=b)?

Do:

$\frac{10+x}{10}$10+x10​	$=$=	$\frac{x+10}{10}$x+1010​	Multiply both sides by $10$10 to remove the denominators
$\frac{10\left(10+x\right)}{10}$10(10+x)10​	$=$=	$\frac{10\left(x+10\right)}{10}$10(x+10)10​	Simplify
$10+x$10+x	$=$=	$x+10$x+10

Can you see that the LHS of this equation is the same as the RHS? They are equivalent expressions but the terms are just switched around. This means this equation has an infinite number of solutions.

 

Question 3

Form an equation to represent the relationship, "three more than three times a number is equal to six more than four times the number".

a) Let $n$n be the number. Note: Do not solve the equation.

Think: We need to split this sentence at the word "equal" to make each side of the equation.

Do:

How would we write "three more than three times a number" mathematically? $3n+3$3n+3.

Similarly, to write "six more than four times the number" mathematically, we would write $4n+6$4n+6.

So, the final equation is $3n+3=4n+6$3n+3=4n+6.

 

b) How many numbers satisfy the relationship? In other words, how many solutions does the equation have?

Think: If we solved this equation, what form would it be in?

Do: The solution to this equation in $n=-3$n=−3, so only one number satisfies this relationship.

 

Question 4

How many solutions does $x=x-10$x=x−10 have?

Think: Can a number ever equal the same number minus $10$10?

Do: This equation is in the form $a=b$a=b because a number can never equal the same number minus $10$10. Therefore, there are no solutions to this equation.

Another way to see that there are no solutions is to complete the equation solving process.

$x$x	$=$=	$x-10$x−10
$x-x$x−x	$=$=	$x-10-x$x−10−x	subtract x from both sides
$0$0	$=$=	$-10$−10	because $0$0 is not equal to $-10$−10, we now know there are no solutions.

 

Question 5