Sec 2 Maths: Simultaneous Linear Equations, practice questions & worked solutions

From (1), $x=20-y$ —(3). Substitute (3) into (2):

$$\begin{aligned}2(20-y)+3y&=48\\ 40-2y+3y&=48\\ y&=8\end{aligned}$$

Substitute $y=8$ into (3):

$$\begin{aligned}x&=20-8\\ &=12\end{aligned}$$

So $x=12$, $y=8$. Check in (2): $2(12)+3(8)=24+24=48$.

By substitution

From (1), $y=34-8x$ —(3). Substitute (3) into (2):

$$\begin{aligned}3x-2(34-8x)&=8\\ 3x-68+16x&=8\\ 19x&=76\\ x&=4\end{aligned}$$

Substitute $x=4$ into (3):

$$\begin{aligned}y&=34-8(4)\\ &=2\end{aligned}$$

By elimination

(1) $\times2$ gives $16x+2y=68$ —(4). The $y$ terms have opposite signs, so add (4) and (2):

$$\begin{aligned}16x+3x&=68+8\\ 19x&=76\\ x&=4\end{aligned}$$

Then $8(4)+y=34$, so $y=2$. Either way, $x=4$, $y=2$.

From (1):

$$\begin{aligned}2y&=3x-8\\ y&=\frac{3x-8}{2}\quad\text{—(3)}\end{aligned}$$

Substitute (3) into (2), then multiply through by $2$:

$$\begin{aligned}4x+5\left(\frac{3x-8}{2}\right)&=3\\ 8x+15x-40&=6\\ 23x&=46\\ x&=2\end{aligned}$$

Substitute $x=2$ into (3):

$$\begin{aligned}y&=\frac{3(2)-8}{2}\\ &=-1\end{aligned}$$

So $x=2$, $y=-1$. Check in (2): $4(2)+5(-1)=8-5=3$.

Make the $y$ coefficients equal: (1) $\times2$ and (2) $\times3$:

$$\begin{aligned}8x-6y&=-14\quad\text{—(3)}\\ 45x-6y&=60\quad\text{—(4)}\end{aligned}$$

The $y$ terms have the same sign, so subtract: (4) $–$ (3):

$$\begin{aligned}45x-8x&=60-(-14)\\ 37x&=74\\ x&=2\end{aligned}$$

Substitute $x=2$ into (1):

$$\begin{aligned}4(2)-3y&=-7\\ -3y&=-15\\ y&=5\end{aligned}$$

So $x=2$, $y=5$. Check in (2): $15(2)-2(5)=30-10=20$.

(a)

Rewrite both equations in the form $y=mx+c$:

$$\begin{aligned}y&=-4x+8\\ y&=-4x+20\end{aligned}$$

Both lines have gradient $-4$ but different $y$-intercepts ($8$ and $20$), so they are parallel and never meet. The equations have no solution (0 solutions).

(b)

$y=4$ is the horizontal line through $4$ on the $y$-axis:

(c)

The line $y=4$ crosses $4x+y=8$ at

$$\begin{aligned}(1,\,4)\end{aligned}$$

Check: $4(1)+4=8$.

(a)

$$\begin{aligned}4(2x+y)&=36\\ 2x+y&=9\\ y&=-2x+9\end{aligned}$$

(b)

The breadth of $A$ is $4$ cm, so the side of $B$ is $4+2=6$ cm:

$$\begin{aligned}x-y&=6\\ y&=x-6\end{aligned}$$

(c)

Both expressions equal $y$, so set them equal:

$$\begin{aligned}-2x+9&=x-6\\ -3x&=-15\\ x&=5\end{aligned}$$
$$\begin{aligned}y&=5-6\\ &=-1\end{aligned}$$

So $x=5$, $y=-1$. Check: $2x+y=10-1=9$ and $x-y=5+1=6$.

(d)

$$\begin{aligned}\text{Area of }A&=36\text{ cm}^2\\ \text{Area of }B&=(x-y)^2\\ &=6^2\\ &=36\text{ cm}^2\end{aligned}$$

I agree with David. Rectangle $A$ gives $36$ squares of side $1$ cm, and square $B$ has area $36$ cm$^2$, so the $36$ squares fit exactly as a $6$ by $6$ arrangement covering $B$.

Let the original fraction be $\dfrac{x}{y}$. Cross-multiply each condition:

$$\begin{aligned}\frac{x-1}{y+3}&=\frac16\\ 6x-6&=y+3\\ y&=6x-9\quad\text{—(1)}\end{aligned}$$
$$\begin{aligned}\frac{x+2}{y-5}&=\frac35\\ 5x+10&=3y-15\\ 3y&=5x+25\quad\text{—(2)}\end{aligned}$$

Substitute (1) into (2):

$$\begin{aligned}3(6x-9)&=5x+25\\ 18x-27&=5x+25\\ 13x&=52\\ x&=4\end{aligned}$$
$$\begin{aligned}y&=6(4)-9\\ &=15\end{aligned}$$

The original fraction is $\dfrac{4}{15}$.

Check: $\dfrac{4-1}{15+3}=\dfrac{3}{18}=\dfrac16$ and $\dfrac{4+2}{15-5}=\dfrac{6}{10}=\dfrac35$.