Worked examples · Sheet 1

Equations of the form |A| = k and |A| = |B|

Theory used: Equation of the form |A| = k Equation of the form |A| = |B|

Exercise 1

Solve the equation

∣2 x - 3∣ = 5.

Since $5 > 0$ , the equation is equivalent to the union

2 x - 3 = 5 \lor 2 x - 3 = - 5.

2 x = 8 \lor 2 x = - 2

x = 4 \lor x = - 1

The solutions are $x = - 1$ and $x = 4$ .

The right-hand side is the constant $5$ , which is positive, so the rule for $∣ A ∣ = k$ with $k > 0$ applies: the expression $2 x - 3$ equals $5$ or $- 5$ . Each of the two equations is linear and is solved by moving the constant term to the right and dividing by $2$ . Checking by substitution confirms the result: for $x = 4$ we have $∣8 - 3∣ = 5$ , and for $x = - 1$ we have $∣ - 2 - 3∣ = 5$ .

Exercise 2

Solve the equations

∣4 x - 1∣ = 0 and ∣3 x + 2∣ = - 5.

First equation: the only number with absolute value zero is $0$ , so

4 x - 1 = 0

x = \frac{1}{4}

Second equation: for every $x \in R$ ,

∣3 x + 2∣ \geq 0 > - 5,

so the equation has no solutions.

The first equation has solution $x = \frac{1}{4}$ ; the second has no solutions.

In the first equation the right-hand side is $0$ , so the absolute value vanishes only where its argument vanishes, and the equation becomes $4 x - 1 = 0$ . In the second the right-hand side is negative, while the left-hand side is an absolute value and is therefore never less than $0$ ; since no value of $x$ makes the two sides equal, the solution set is empty.

Exercise 3

Solve the equation

∣ x - 1∣ = ∣2 x + 3∣.

Two numbers have the same absolute value when they are equal or opposite, so

x - 1 = 2 x + 3 \lor x - 1 = - (2 x + 3) .

- x = 4 \lor 3 x = - 2

x = - 4 \lor x = - \frac{2}{3}

The solutions are $x = - 4$ and $x = - \frac{2}{3}$ .

Both sides are absolute values, so neither of them can be negative and no acceptability condition is needed: the equation is equivalent to the union of the two cases, equal arguments or opposite arguments. In the second case the minus sign must be distributed over the whole second argument, therefore $x - 1 = - 2 x - 3$ . Checking by substitution confirms the result: for $x = - 4$ both sides equal $∣ - 5∣ = 5$ , and for $x = - \frac{2}{3}$ both equal $\frac{5}{3}$ .