Systems of Linear Equations

HKC78-AIHL-SYS-01

Paper 1 Extended Response

Medium No Calculator

Solve

$3x + y = 10$,

$x + 2y – z = 1$,

$2x – z = 4$.

Question HKC78-AIHL-SYS-01 ×

Question HKC78-AIHL-SYS-01

[Maximum mark: 6]

Solve

$3x + y = 10$,

$x + 2y – z = 1$,

$2x – z = 4$.

Mark Scheme

From the third equation, $z = 2x – 4$. From the first, $y = 10 – 3x$. [1]

Substitute into the second: $x + 2(10 – 3x) – (2x – 4) = 1$. [2]

$x + 20 – 6x – 2x + 4 = 1$, so $-7x + 24 = 1$ and $x = dfrac{23}{7}$. [2]

$y = 10 – dfrac{69}{7} = dfrac{1}{7}$ and $z = dfrac{46}{7} – 4 = dfrac{18}{7}$. [1]

Answer: $x = dfrac{23}{7}$, $y = dfrac{1}{7}$, $z = dfrac{18}{7}$.

Total: 6 marks

HKC78-AIHL-SYS-02

Paper 1 Extended Response

Easy No Calculator

Four notebooks and six pens cost $\$38$. Three notebooks and two pens cost $\$20$. Find the price of one notebook and the price of one pen.

Question HKC78-AIHL-SYS-02 ×

Question HKC78-AIHL-SYS-02

[Maximum mark: 5]

Four notebooks and six pens cost $\$38$. Three notebooks and two pens cost $\$20$. Find the price of one notebook and the price of one pen.

Mark Scheme

Let $n$ be the notebook price and $p$ the pen price. Then $4n + 6p = 38$ and $3n + 2p = 20$. Divide the first equation by $2$: $2n + 3p = 19$. [1]

Multiply $2n + 3p = 19$ by $3$ and the second equation by $2$: $6n + 9p = 57$ and $6n + 4p = 40$. [1]

Subtract: $5p = 17$, so $p = dfrac{17}{5} = 3.4$. [2]

$3n + 2(3.4) = 20$, so $3n + 6.8 = 20$, $3n = 13.2$ and $n = 4.4$. A notebook costs $$4.40$ and a pen costs $$3.40$. [1]

Total: 5 marks