Inequalities

New Zealand

Level 6 - NCEA Level 1

Lesson

In most cases when we have an inequality involving a variable, it will be useful to simplify and solve the inequality.

We are used to solving equations, and the method for solving inequalities is very similar. Most importantly, "whatever operation is done to one side needs to be done to the other side" is still true for inequalities!

In fact, the steps that we take are the **same steps** as for equations, but with one extra consideration: **which way does the inequality symbol face after each step**?

Addition:

Numeric | Algebraic | |||||

$3$3 | $<$< | $4$4 | $x-2$x−2 |
$<$< | $4$4 | |

$3+2$3+2 | $<$< | $4+2$4+2 | $x-2+2$x−2+2 |
$<$< | $4+2$4+2 | |

$5$5 | $<$< | $6$6 | $x$x |
$<$< | $6$6 |

As we can see, adding a number to both sides of an inequality **doesn't change** the inequality symbol.

Subtraction:

Numeric | Algebraic | |||||

$5$5 | $>$> | $2$2 | $x+1$x+1 |
$\ge$≥ | $2$2 | |

$5-1$5−1 | $>$> | $2-1$2−1 | $x+1-1$x+1−1 |
$\ge$≥ | $2-1$2−1 | |

$4$4 | $>$> | $1$1 | $x$x |
$\ge$≥ | $1$1 |

Similarly, subtracting a number from both sides of an inequality **doesn't change** the inequality symbol.

Multiplication by positive numbers:

Numeric | Algebraic | |||||

$4$4 | $<$< | $7$7 | $\frac{x}{3}$x3 |
$\le$≤ | $7$7 | |

$4\times3$4×3 | $<$< | $7\times3$7×3 | $\frac{x}{3}\times3$x3×3 |
$\le$≤ | $7\times3$7×3 | |

$12$12 | $<$< | $21$21 | $x$x |
$\le$≤ | $21$21 |

Multiplication by negative numbers:

Numeric | Algebraic | |||||

$4$4 | $<$< | $7$7 | $-\frac{x}{3}$−x3 |
$<$< | $7$7 | |

$4\times\left(-3\right)$4×(−3) | $>$> | $7\times\left(-3\right)$7×(−3) | $\left(-\frac{x}{3}\right)\times\left(-3\right)$(−x3)×(−3) |
$>$> | $7\times\left(-3\right)$7×(−3) | |

$-12$−12 | $>$> | $-21$−21 | $x$x |
$>$> | $-21$−21 |

Now we have found a difference! Multiplying both sides of an inequality by a **positive** number **doesn't change** the inequality symbol, but multiplying by a **negative** number **does** change the inequality symbol.

Division:

Remember that we can think about division as an equivalent multiplication. For example, dividing by $3$3 is the same as multiplying by $\frac{1}{3}$13. So we get the same result: dividing both sides of an inequality by a **positive** number **doesn't change** the inequality symbol, but dividing by a **negative** number **does** change the inequality symbol.

It is often useful to write an equation in reverse order. For example, if we reach the step $3=x$3=`x` in our working, we often reverse this and write $x=3$`x`=3.

We can also do this with inequalities, as long as we **reverse the symbol** as well. For example, "$3$3 is greater than $x$`x`" means the same thing as "$x$`x` is less than $3$3". So $3>x$3>`x` can be rewritten as $x<3$`x`<3, using the opposite symbol.

Same symbol

The following operations don't change the inequality symbol used:

**Adding**a number to both sides of an inequality.**Subtracting**a number from both sides of an inequality.**Multiplying**both sides of an inequality by a**positive**number.**Dividing**both sides of an inequality by a**positive**number.

Opposite symbol

The following operations reverse the inequality symbol used:

- Writing an inequality in reverse order.
**Multiplying**both sides of an inequality by a**negative**number.**Dividing**both sides of an inequality by a**negative**number.

Solve the following inequality: $x+5>14$`x`+5>14

Solve the following inequality: $x+5\ge11$`x`+5≥11

Solve the following inequality: $10x<90$10`x`<90

Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns

Apply algebraic procedures in solving problems