Algebra I

Hong Kong

Stage 1 - Stage 3

Lesson

In algebra we are very often required to substitute specific values into variables and constants.

The great philosopher and mathematician Rene Descartes (1596-1650) decided to use letters at the front of the alphabet for constants and letters at the back of the alphabet for variables. Generally speaking, this convention is still in use today.

You may recall the equation of the straight line as $y=mx+b$`y`=`m``x`+`b` where $m$`m` is the gradient and $b$`b` is the y intercept.

The constants in this equation are the gradient $m$`m` and the $y$`y` intercept $b$`b`. If we give them specific values then we identify a specific line. For example, if $m=2$`m`=2 and $b=5$`b`=5 then the specific line becomes $y=2x+5$`y`=2`x`+5.

The variables $x$`x` and $y$`y` become the coordinates on that line. We can freely choose $x$`x` to be anything we want it to be, and our choice will determine $y$`y`.

For example, if $x=7$`x`=7, then by substitution $y=2\times7+5$`y`=2×7+5. This means $y$`y` becomes $19$19. The point $(7,19)$(7,19) is on the line.

What's important here is that any formula you meet contains letters and, when the formula is applied to a specific situation, specific numbers are substituted to evaluate specific answers.

There are groups of useful formulae. We will explore some examples here.

The arithmetic average $M$`M` of two numbers $a$`a` and $b$`b` is given by $M=\frac{a+b}{2}$`M`=`a`+`b`2. Evaluate $M$`M` if $a=23$`a`=23 and $b=77$`b`=77.

Then $M=\frac{23+77}{2}=\frac{100}{2}=50$`M`=23+772=1002=50

The temperature $F$`F` in degrees Fahrenheit is found using the formula $F=\frac{9}{5}C+32$`F`=95`C`+32 where $C$`C` is the temperature in degrees Celsius. Find $F$`F` when $C=100$`C`=100.

Then $F=\frac{9}{5}\left(100\right)+32=212$`F`=95(100)+32=212.

A certain quantity $H$`H` is determined by the formula $H=\frac{2}{\frac{1}{u}+\frac{1}{v}}$`H`=21`u`+1`v`. Find $H$`H` when $u=9$`u`=9 and $v=18$`v`=18.

This looks complicated, so we will proceed step by step;

$H$H |
$=$= | $\frac{2}{\frac{1}{u}+\frac{1}{v}}$21u+1v |

$=$= | $\frac{2}{\frac{1}{9}+\frac{1}{18}}$219+118 | |

$=$= | $\frac{2}{\frac{2}{18}+\frac{1}{18}}$2218+118 | |

$=$= | $\frac{2}{\left(\frac{3}{18}\right)}$2(318) | |

$=$= | $2\times\frac{18}{3}$2×183 | |

$=$= | $2\times6=12$2×6=12 |

Substitute $x=3$`x`=3 and $y=-4$`y`=−4 into the expression $\frac{x^2}{3}+\frac{y^3}{8}-xy$`x`23+`y`38−`x``y`.

Then the expression evaluates as:

$\frac{3^2}{3}+\frac{\left(-4\right)^3}{8}-\left(3\right)\left(-4\right)=3-8+12=7$323+(−4)38−(3)(−4)=3−8+12=7.

Ignoring air resistance, a rock dropped off the edge of a cliff will fall through the air distance of $d$`d` metres in $t$`t` seconds where $d=4.9\times t^2$`d`=4.9×`t`2.

Assuming it hasn't hit the ground after $3$3 seconds, how far did it fall in the third second.

To answer this, we need to substitute twice into the formula.

Over $2$2 seconds, the formula shows that the rock will fall $d=4.9\times2^2=19.6$`d`=4.9×22=19.6 metres

Over $3$3 seconds, the formula shows that the rock will fall $d=4.9\times3^2=44.1$`d`=4.9×32=44.1 metres

So therefore the rock falls $44.1-19.6=24.5$44.1−19.6=24.5 metres in the third second. Tricky!

Evaluate $2v$2`v` when $v=-3$`v`=−3.

Evaluate $2m\times3n+37$2`m`×3`n`+37 when $m=4$`m`=4 and $n=5$`n`=5.

Evaluate $\frac{6}{7x+3}$67`x`+3 when $x=3$`x`=3. Make sure your answer is fully simplified.