The pair (x, y) satisfying x + y = 5; 2x - y = 1 is:

Adding: 3x = 6 ⇒ x = 2. Then y = 5 - 2 = 3. So (x, y) = (2, 3).

For which value of a ∈ ℝ does the graph of f(x) = ax + 3 pass through the point (2, 7)?

f(2) = 2a + 3 = 7 ⇒ 2a = 4 ⇒ a = 2.

The solution set of 2x - 3 > 5 in ℝ is:

2x > 8 ⇒ x > 4, so the set is (4, +∞).

Solve -2x + 5 ≤ 11.

-2x + 5 ≤ 11 -2x ≤ 6 Divide both sides by -2 (flip the sign): x ≥ -3.

IB Maths AA

Practice by topic

Linear functions and models practice questions — IB Maths AA

12 free multiple-choice problems on linear functions and models, ordered to match IB Maths AA difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

Try 3 problems from this chapter

1 / 3

Problem #0321 International

Beginnerlinear-function

For,, the value of for which is:

Problems & worked solutions

Problem #0321 International

Problem 1 — Linear functions and models

For $f : R \to R$ , $f (x) = 2 x - 4$ , the value of $x$ for which $f (x) = 0$ is:

Show answer & worked solution

A. $- 4$
B. $- 2$
C. $2$ ✓ correct
D. $4$

$2 x - 4 = 0 \Rightarrow x = 2$ .

Problem #0322 International

Problem 2 — Linear functions and models

The solution of $3 x + 5 = 14$ is:

Show answer & worked solution

A. $2$
B. $3$ ✓ correct
C. $4$
D. $\frac{14}{3}$

$3 x = 14 - 5 = 9 \Rightarrow x = 3$ .

Problem #0323 International

Problem 3 — Linear functions and models

Solve $2 (x - 1) = 4 x + 6$ .

Show answer & worked solution

A. $- 4$ ✓ correct
B. $- 2$
C. $2$
D. $4$

$2 x - 2 = 4 x + 6 \Rightarrow - 2 x = 8 \Rightarrow x = - 4$ .

Problem #0325 International

Problem 4 — Linear functions and models

The pair $(x, y)$ satisfying ${\begin{cases} x + y = 5 \\ 2 x - y = 1 \end{cases}$ is:

Show answer & worked solution

A. $(1, 4)$
B. $(2, 3)$ ✓ correct
C. $(3, 2)$
D. $(4, 1)$

Adding: $3 x = 6 \Rightarrow x = 2$ . Then $y = 5 - 2 = 3$ . So $(x, y) = (2, 3)$ .

Problem #0328 International

Problem 5 — Linear functions and models

For which value of $a \in R$ does the graph of $f (x) = a x + 3$ pass through the point $(2, 7)$ ?

Show answer & worked solution

A. $1$
B. $2$ ✓ correct
C. $3$
D. $7$

$f (2) = 2 a + 3 = 7 \Rightarrow 2 a = 4 \Rightarrow a = 2$ .

Problem #0910 US SAT

Problem 6 — Linear functions and models

The solution to $3 x - 7 = 2 x + 5$ is:

Show answer & worked solution

A. $x = - 2$
B. $x = 2$
C. $x = 12$ ✓ correct
D. $x = - 12$

$3 x - 7 = 2 x + 5$

$3 x - 2 x = 5 + 7$

$x = 12$ .

Problem #0324 International

Problem 7 — Linear functions and models

The solution set of $2 x - 3 > 5$ in $R$ is:

Show answer & worked solution

A. $(- \infty, 4)$
B. $(- \infty, 4]$
C. $(4, + \infty)$ ✓ correct
D. $[4, + \infty)$

$2 x > 8 \Rightarrow x > 4$ , so the set is $(4, + \infty)$ .

Problem #0326 International

Problem 8 — Linear functions and models

The slope of the line through $A (1, 2)$ and $B (4, 11)$ is:

Show answer & worked solution

A. $\frac{1}{3}$
B. $2$
C. $3$ ✓ correct
D. $9$

$m = \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3$ .

Problem #0911 US SAT

Problem 9 — Linear functions and models

Solve $- 2 x + 5 \leq 11$ .

Show answer & worked solution

A. $x \geq - 3$ ✓ correct
B. $x \leq - 3$
C. $x \geq 3$
D. $x \leq 3$

$- 2 x + 5 \leq 11$

$- 2 x \leq 6$

Divide both sides by $- 2$ (flip the sign): $x \geq - 3$ .

Problem #0327 International

Problem 10 — Linear functions and models

The solution set of $∣ 2 x - 1 ∣ = 5$ is:

Show answer & worked solution

A. ${3}$
B. ${- 2}$
C. ${- 2, 3}$ ✓ correct
D. ${2, - 3}$

$2 x - 1 = 5 \Rightarrow x = 3$ , or $2 x - 1 = - 5 \Rightarrow x = - 2$ . Solution set: ${- 2, 3}$ .

2 more Linear functions and models questions in the app

Also covered in Linear functions and models practice across every exam.

Not sure where you stand? Take the free 10-question placement test — no account, ~15 minutes.