MYP 3 · Maths
LAWS OF ALGEBRA
The distributive law
QUESTION 1 [5 marks] — Criterion A
Medium
Expand each expression:
a.
$4(x+6)$
[1] Show Solution
$$4x+24$$
b.
$-3(2x-5)$
[1] Show Solution
$$-6x+15$$
c.
$x(x+7)$
[1] Show Solution
$$x^2+7x$$
d.
$-2x(3x-4)$
[2] Show Solution
$$-6x^2+8x$$
QUESTION 2 [4 marks] — Criterion B
Medium
Investigate why $a(b+c)=ab+ac$ is true, using areas of rectangles.
a.
A rectangle has width $a$ and length $(b+c)$, split into two smaller rectangles of length $b$ and $c$. Write an expression for the area of the WHOLE rectangle two ways: (i) directly as width times total length, and (ii) as the sum of the two smaller rectangles' areas.
[3] Show Solution
(i) $a(b+c)$. (ii) $ab + ac$ (the two smaller rectangle areas added together).
b.
Since both expressions describe the exact same total area, what does this prove about the distributive law?
[1] Show Solution
Since both must be equal (same total area, calculated two ways), this proves $a(b+c)=ab+ac$.
QUESTION 3 [4 marks] — Criterion C
Medium
A student expands $5(2x+3)$ as $10x+3$ (forgetting to multiply the second term by 5).
a.
Explain the student's error, and give the correct expansion.
[2] Show Solution
The student only multiplied the FIRST term inside the brackets by 5, forgetting that the 5 must multiply EVERY term inside the brackets. Correct: $5(2x+3)=10x+15$.
b.
Verify the correct answer by substituting $x=2$ into both the original expression $5(2x+3)$ and your expanded answer.
[2] Show Solution
Original: $5(2(2)+3)=5(7)=35$. Expanded: $10(2)+15=20+15=35$ ? — matches, confirming the correct expansion.
QUESTION 4 [4 marks] — Criterion D
Medium
A cinema charges $x$ dollars per adult ticket. A family buys 3 adult tickets and also pays a \$4 booking fee per ticket.
a.
Write an expression for the total cost, in the form $3(x+4)$, and expand it.
[2] Show Solution
$$3(x+4)=3x+12$$
b.
If tickets cost $x=\$12$ each, find the total cost using both the unexpanded and expanded forms, confirming they match.
[2] Show Solution
Unexpanded: $3(12+4)=3(16)=48$. Expanded: $3(12)+12=36+12=48$ ?.
QUESTION 5 [2 marks] — Criterion A
Easy
Expand: $$-4(3x-2)$$
a.
Expand carefully, paying attention to the negative sign outside the bracket.
[2] Show Solution
$$-4(3x-2) = -12x+8$$
QUESTION 6 [5 marks] — Criterion B
Medium
Investigate the distributive law applied to THREE terms inside the bracket, e.g. $a(b+c+d)$, using $3(x+y+2)$.
a.
Expand $3(x+y+2)$, then verify by substituting $x=4,y=5$ into BOTH the original and expanded form.
[3] Show Solution
$3(x+y+2)=3x+3y+6$. Original at $x=4,y=5$: $3(4+5+2)=3(11)=33$. Expanded: $3(4)+3(5)+6=12+15+6=33$ ?.
b.
State the general rule: does the distributive law work the same way for ANY number of terms inside the bracket (not just 2), and explain briefly why.
[2] Show Solution
Yes — the distributive law extends to any number of terms: $a(b_1+b_2+\cdots+b_n)=ab_1+ab_2+\cdots+ab_n$. This works because multiplication distributes over EACH addition individually, regardless of how many terms are being added together inside the bracket.
QUESTION 7 [5 marks] — Criterion C
Medium
A student simplifies $x(x+7)-3(x-2)$ as $x^2+7x-3x-2$ (forgetting to multiply the $-3$ by BOTH terms in the second bracket).
a.
Explain the error precisely, and give the fully correct expansion.
[3] Show Solution
The student correctly distributed $x(x+7)=x^2+7x$, but only multiplied $-3$ by the $x$ term (getting $-3x$) and forgot to multiply $-3$ by $-2$ as well. Correct: $x(x+7)-3(x-2)=x^2+7x-3x+6=x^2+4x+6$.
b.
Verify the correct answer using $x=3$.
[2] Show Solution
Original: $3(10)-3(1)=30-3=27$. Correct formula $x^2+4x+6$ at $x=3$: $9+12+6=27$ ?.
QUESTION 8 [3 marks] — Criterion C
Medium
A classmate says the distributive law $a(b+c)=ab+ac$ 'only works for addition, never subtraction', after seeing it written that way in a textbook.
a.
Explain why the distributive law ALSO applies to subtraction (i.e. $a(b-c)=ab-ac$), by rewriting subtraction as 'adding a negative' and applying the SAME rule.
[3] Show Solution
Subtraction can always be rewritten as addition of a negative: $b-c=b+(-c)$. Applying the distributive law to THIS form: $a(b+(-c))=ab+a(-c)=ab-ac$ — showing that the subtraction version is really just a special case of the addition version, using a negative term, so the SAME underlying rule covers both.