Sec 2 Maths: Linear Inequalities, practice questions & worked solutions

Divide both sides by $-2$ and reverse the sign:

$$\begin{aligned}x&\leq12\end{aligned}$$

(a)

$$\begin{aligned}2x+5&\gt8\\ 2x&\gt3\\ x&\gt1.5\end{aligned}$$

(b)

$x$ must be greater than $1.5$, and $1.5$ itself is not allowed. The smallest integer is

$$\begin{aligned}x&=2\end{aligned}$$

(a)

$$\begin{aligned}5-2x&\geq3x+10\\ -2x-3x&\geq10-5\\ -5x&\geq5\end{aligned}$$

Divide both sides by $-5$ and reverse the sign:

$$\begin{aligned}x&\leq-1\end{aligned}$$

(b)

Draw a solid circle at $-1$, because $-1$ is included, with an arrow pointing to the left:

(a)

$$\begin{aligned}2y-1&\geq3(y-3)\\ 2y-1&\geq3y-9\\ -y&\geq-8\end{aligned}$$

Multiply both sides by $-1$ and reverse the sign:

$$\begin{aligned}y&\leq8\end{aligned}$$

(b)

Draw a solid circle at $8$, because $8$ is included, with an arrow pointing to the left:

(c)

The primes up to $8$ are $2$, $3$, $5$ and $7$. The largest is

$$\begin{aligned}7\end{aligned}$$

(a)

Multiply both sides by $2$ to clear the fraction:

$$\begin{aligned}2(x-3)&\lt5x\\ 2x-6&\lt5x\\ -3x&\lt6\end{aligned}$$

Divide both sides by $-3$ and reverse the sign:

$$\begin{aligned}x&\gt-2\end{aligned}$$

(b)

Draw a hollow circle at $-2$, because $-2$ is not included, with an arrow pointing to the right:

(c)

$x$ must be greater than $-2$, so $-2$ is excluded. The smallest integer is

$$\begin{aligned}x&=-1\end{aligned}$$

(a)

$$\begin{aligned}5-x&\lt\frac{11}{2}\\ -x&\lt\frac{11}{2}-5\\ -x&\lt\frac12\end{aligned}$$

Multiply both sides by $-1$ and reverse the sign:

$$\begin{aligned}x&\gt-\frac12\end{aligned}$$

(b)

Every prime number is greater than $-\dfrac12$, so the smallest prime that satisfies the inequality is the smallest prime of all:

$$\begin{aligned}2\end{aligned}$$

First solve the inequality. Multiply both sides by $2$:

$$\begin{aligned}\frac{3(x-1)}{2}&\leq x+1\\ 3x-3&\leq2x+2\\ x&\leq5\end{aligned}$$

(a)

The perfect squares up to $5$ are $0$, $1$ and $4$. The greatest is

$$\begin{aligned}x&=4\end{aligned}$$

(b)

$5$ is included (the sign is $\leq$) and $5$ is prime, so the greatest prime is

$$\begin{aligned}x&=5\end{aligned}$$

(a)(i)

Multiply every term by the LCM of the denominators, $20$:

$$\begin{aligned}4x+5(2x+3)&=120\\ 4x+10x+15&=120\\ 14x&=105\\ x&=7\tfrac12\end{aligned}$$

(a)(ii)

Bring every term to one side and factorise:

$$\begin{aligned}5x^2-27x+10&=0\\ (5x-2)(x-5)&=0\\ x=\frac25\quad&\text{or}\quad x=5\end{aligned}$$

(b)

$$\begin{aligned}5(4x-7)&\geq7(1+3x)\\ 20x-35&\geq7+21x\\ 20x-21x&\geq7+35\\ -x&\geq42\end{aligned}$$

Multiply both sides by $-1$ and reverse the sign:

$$\begin{aligned}x&\leq-42\end{aligned}$$

There are $x$ notes of $20, so there are $(12-x)$ notes of $5. Work in dollars: $16\,000$ cents $=$ $160.

$$\begin{aligned}20x+5(12-x)&\lt160\\ 20x+60-5x&\lt160\\ 15x&\lt100\\ x&\lt6\tfrac23\end{aligned}$$

$x$ is a whole number of notes less than $6\tfrac23$, so the maximum number of $20 notes is $6$.

Check: $6$ notes of $20 and $6$ of $5 total $150, which is less than $160; $7$ and $5$ would total $165.