Equations

Lesson

In this set of questions, we will combine many of our skills with equations and graphs to solve a wide range of worded problems. We will be looking at problems that involve:

- Quadratic Equations, e.g. $y=3x^2+4x-18$
`y`=3`x`2+4`x`−18 - Cubic Equations, e.g. $y=8x^3-3$
`y`=8`x`3−3 - Radical Equations, e.g. $y=\sqrt{x-3}+5$
`y`=√`x`−3+5 - Rational Equations, e.g. $y=\frac{15}{x-3}$
`y`=15`x`−3 - Exponential Equations, e.g. $y=285\left(1.7\right)^x$
`y`=285(1.7)`x`

Let's say you had a friend named Joe, who was a Zookeeper. He wants to build an enclosure, and only has enough money in his budget to buy $22$22 metres of fencing. He wants to get as many animals into the enclosure as possible. This means we are looking we are looking to maximise the area of the enclosure with a fixed perimeter of $22$22 m.

The following applet models an enclosure with $22$22m of fencing. You will notice that the side lengths, perimeter, and area, are calculated, and point is graphed to show the change in area as the side lengths change. Drag point D, the green point, to see it all change. By playing with the Geogebra Applet, you will be able to get a good idea of what the maximum area will be. Explore with the applet and see if you can find the maximum area.

Let's now explore how we can solve this algebraically.

We are attempting to maximise the area of the enclosure. The area of the enclosure is:

$\text{area}=\text{length}\times\text{width}$area=length×width

This means we have three variables, $\text{area}$area, $\text{length}$length, and $\text{width}$width. As we don't usually work with three variables in this way, we want to find any other relationships we can, then substitute that relationship into our $\text{area}$area formula, and use our skills to maximise the area of the enclosure.

Let's look at the fencing around the enclosure. The length of the fencing around the enclosure relates to the perimeter of the rectangular closure.

This gives us the expression:

$2\left(\text{length}+\text{width}\right)=22$2(length+width)=22

Dividing both sides by $2$2, and making $\text{length}$length the subject, we get:

$\text{length}=11-\text{width}$length=11−width

To make the manipulation easier, we will let: $\text{width}=w$width=`w`.

$\therefore\text{length}=11-w$∴length=11−`w`

Now, we can substitute that expression into the area formula to get:

$A=w\left(11-w\right)$`A`=`w`(11−`w`)

Our goal is to maximise the area. As we saw in the graph above, the area is $0$0 when when the width is $0$0 or $11$11. As it is a parabola, the maximum area occurs in the middle of these two values, when $w=5.5$`w`=5.5.

If we substitute w=5.5 into the area formula, we get:

$A=w\left(11-w\right)$`A`=`w`(11−`w`)$=$=$5.5\left(11-5.5\right)=30.25$5.5(11−5.5)=30.25 m^{2}.

Therefore, the maximum area of the enclosure is $30.25$30.25 m^{2}.

A cube has side length $7$7 cm and a mass of $1715$1715 g. The mass of the cube is directly proportional to the cube of its side length.

Let $k$

`k`be the constant of proportionality for the relationship between the side length $x$`x`and the mass $m$`m`of a cube. Solve for $k$`k`.Hence state the equation relating the mass ($m$

`m`) and side length ($x$`x`) of a cube.Fill in the following table of values for $m=5x^3$

`m`=5`x`3$x$ `x`1 2 3 4 $m$ `m`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Choose the correct curve for $m=5x^3$

`m`=5`x`3Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DLoading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DFrom the equation, find the mass of a cube with side $8.5$8.5 cm, to the nearest gram.

A cube has a mass of $1920$1920 g. From the graph, determine what whole number value its side length is closest to.

In a laboratory, a drug is tested on a sample of $2880$2880 cancer cells. The number of cancer cells is recorded and the cancer cells are found to halve each week.

Complete the table of values.

number of weeks passed ($x$ `x`)$0$0 $1$1 $2$2 $3$3 $4$4 number of cells ($y$ `y`)$2880$2880 $\editable{}$ $720$720 $\editable{}$ $\editable{}$ Which general equation satisfies this model?

Assume $a$

`a`and $b$`b`are positive integers.$y=ba^{-x}$

`y`=`b``a`−`x`A$y=ba^x$

`y`=`b``a``x`B$y=-a^{-x}$

`y`=−`a`−`x`C$y=a^{-x}$

`y`=`a`−`x`D$y=ba^{-x}$

`y`=`b``a`−`x`A$y=ba^x$

`y`=`b``a``x`B$y=-a^{-x}$

`y`=−`a`−`x`C$y=a^{-x}$

`y`=`a`−`x`DFind the equation linking the number of cancer cells ($y$

`y`) and the number of weeks passed ($x$`x`).According to this model, does the drug completely remove all the cancer cells?

Yes

ANo

BYes

ANo

B

Sophia is considering two different career options. She can either accept a job with a company that has offered her a salary of $\$125000$$125000 a year with a projected salary increase of $\$3000$$3000 per year, or start her own business where she estimates her initial annual salary to be $\$20000$$20000 per year. She projects that if she starts her own business, her annual salary will increase by $40%$40% each year.

Form an equation for $f\left(n\right)$

`f`(`n`), her salary in $n$`n`years time if she accepts the job with a company.Form an equation for $g\left(n\right)$

`g`(`n`), her salary in $n$`n`years time if she starts her own business.Complete the table of values.

$n$ `n`Increase in salary compared to previous year when working for a company Increase in salary compared to previous year when working for her own business $2$2 $\editable{}$ $\editable{}$ $3$3 $\editable{}$ $\editable{}$ $4$4 $\editable{}$ $\editable{}$ After $3$3 years, which option will result in a higher annual salary for Sophia?

working for her own business

Aworking for a company

Bworking for her own business

Aworking for a company

BAfter $6$6 years, which option will result in a higher annual salary for Sophia?

working for a company

Aworking for her own business

Bworking for a company

Aworking for her own business

B

Form and use linear, quadratic, and simple trigonometric equations

Apply graphical methods in solving problems

Apply algebraic methods in solving problems