Algebra I

Hong Kong

Stage 1 - Stage 3

Lesson

In algebra, we use letters to represent unknown values, and we call these unknowns variables or pronumerals. For example if we know that $x+2=5$`x`+2=5, then we can work out that $x=3$`x`=3 since we also know that $3+2=5$3+2=5.

Sometimes we want to do this process in reverse, however, and we substitute numbers into equations in place of variables to determine a final value. We can substitute in any kinds of numbers, including whole numbers, decimals and fractions.

Evaluate**:** If $x=3$`x`=3, evaluate $6x-4$6`x`−4.

Think**: **This means that everywhere the letter $x$`x` has been written, we will replace it with the number $3$3.

Do**:**

$6x-4$6x−4 |
$=$= | $6\times3-4$6×3−4 |

$=$= | $18-4$18−4 | |

$=$= | $14$14 |

The same process applies even if there is more than one unknown value.

Evaluate**: **If $x=6$`x`=6 and $y=0.5$`y`=0.5, evaluate $6x-2y-12$6`x`−2`y`−12.

Think**:** Just like before, we will replace the letter $x$`x` with the number $6$6, and the letter $y$`y` with the number $0.5$0.5. We also need to keep the order of operations in mind when we do these kinds of calculations!

Do**: **

$6x-2y-12$6x−2y−12 |
$=$= | $6\times6-2\times0.5-12$6×6−2×0.5−12 | replacing $x$x with $6$6, and $y$y with $0.5$0.5. |

$=$= | $36-1-12$36−1−12 | evaluating multiplication before subtraction. | |

$=$= | $23$23 |

Now let's watch some worked solutions to the following questions.

Evaluate $8x+4$8`x`+4 when $x=2$`x`=2.

For $x=10$`x`=10 and $y=6$`y`=6, evaluate $8x+6y+4$8`x`+6`y`+4