$textbf{(a)}$
We have given pair of equations as

$$
begin{align}
y&=3x+7\
y&=-4x+21
end{align}
$$

This pair of equation can be solved by substitution method. On putting $y=3x+7$ into the equation (2), it will result in an equation of single variable in terms of $x$ only.

$$
begin{align*}
y&=-4x+21\
(3x+7)&=-4x+21tag{ add $4x$ to the each side}\
3x+4x+7&=-4x+4x+21\
7x+7&=21tag{subtract $7$ from the each side}\
7x+7-7&=21-7\
7x&=14tag{ divide each side by $7$}\
dfrac{7x}{7}&=dfrac{14}{7}\
x&=2
end{align*}
$$

We can solve for $y$ by putting $x=2$ into $y=-4x+21$, which gives

$$
begin{align*}
y&=-4x+21\
y&=-4(2)+21\
y&=-8+21\
y&=13
end{align*}
$$

Therefore , the solution point is $(2,13)$

$textbf{(b)}$
$$
begin{align}
3x-y&=17\
-x+y&=-7
end{align}
$$

We can solve the above system using Elimination method. Combine both the equation of the system and solve as shown below.

$$
begin{align*}
3x-y&=17\
+(-x+y)& +(-7)\
3x-x-y+y&=17-7\
2x&=10tag{ divide each side by $2$}\
dfrac{2x}{2}&=dfrac{10}{2}\
x&=5
end{align*}
$$

Put $x=5$ into the equation $-x+y=-7$ and solve for $y$.

$$
begin{align*}
-x+y&=-7\
-(5)+y&=-7tag{ add $5$ to the each side}\
-5+5+y&=-7+5\
y&=-2
end{align*}
$$

Therefore, the solution point is $(5,-2)$

$textbf{(c)}$
We have given pair of equations as

$$
begin{align}
x&=3y-5\
2x&+12y=-4
end{align}
$$

This pair of equation can be solved by substitution method. On putting $x=3y-5$ into the equation (2), it will result in an equation of single variable in terms of $y$ only.

$$
begin{align*}
2x+12y&=-4\
2(3y-5)+12y&=-4\
6y-10+12y&=-4tag{ add $10$ to the each side}\
18y-10+10&=-4+10\
18y&=6tag{ divide each side by $18$}\
dfrac{18y}{18}&=dfrac{6}{18}\
y&=dfrac{1}{3}
end{align*}
$$

We can solve for $x$ by putting $y=dfrac{1}{3}$ into $x=3y-5$, which gives

$$
begin{align*}
x&=3y-5\
x&=3(dfrac{1}{3})-5\
x&=1-5\
x&=-4
end{align*}
$$

Therefore , the solution point is $(-4,dfrac{1}{3})$

$textbf{(a)}$ The solution point is $(2,13)$

$textbf{(b)}$ The solution point is $(5,-2)$

$textbf{(c)}$ The solution point is $(-4,dfrac{1}{3})$

Let Bob an Roy met after $t$ seconds.

Rate of climb of Bob is two rungs per second downwards

Rate of climb of Roy is one rung per second upwards

Given, that the each ladder has total 30 rungs, then we can write the equations for each case as shown below

$$
begin{align}
text{Rung position of Bob}&=30-2cdot t\
text{Rung position of Roy}&=0+ t
end{align}
$$

We have to find the time when they were on the same height(same rung), hence we have to equate the above equations

$Rightarrow$
$$
begin{align*}
t&=30-2cdot t\
3t&=30\
t&=dfrac{30}{3}\
t&=10
end{align*}
$$

Therefore, they met on the 10th rung.

$$
{color{#4257b2}text{ a) }}
$$

$$
begin{align*}
&text{Add 11 to both sides}\
&6x-11+11=4x+12+11\\
&6x=4x+23 tag{ Simplify} \
&6x-4x=4x+23-4x tag{Subtract 4x from both sides}\
&2x=23 tag{Simplify}\
&frac{2x}{2}=frac{23}{2} tag{Divide both sides by 2}\\
&boxed{{color{#c34632} x= frac{23}{2} }}
end{align*}
$$

$$
{color{#4257b2}text{ b) }}
$$

$$
begin{align*}
&text{Expand}\
&6x-10=6x-4\\
&6x-10+10=6x-4+10 tag{Add 10 to both sides}\
&6x=6x+6 tag{Simplify}\
&6x-6x=6x+6-6x tag{Subtract 6x from both sides}\
&0=-6 tag{Simplify}\\
&boxed{{color{#c34632} text{No solution} }}
end{align*}
$$

$$
{color{#4257b2}text{ c) }}
$$

$$
begin{align*}
&text{Expand}\
&x^2+x-12=x^2+4\\
&x^2+x-12+12=x^2+4+12tag{Add 12 to both sides}\
&x^2+x=x^2+16 tag{Simplify}\
&x^2+x-x^2=x^2+16-x^2 tag{Subtract x2 from both sides}\\
&boxed{{color{#c34632} x=16 }}
end{align*}
$$

$$
{color{#4257b2}text{ d) }}
$$

$$
begin{align*}
&text{Multiply both sides by 10}\
&frac{x}{25}cdot :10=frac{7}{10}cdot :10\\
&frac{2x}{5}=7 tag{Simplify}\
&frac{5cdot :2x}{5}=7cdot :5tag{Multiply both sides by 5}\
&2x=35 tag{Simplify}\
&frac{2x}{2}=frac{35}{2} tag{Divide both sides by 2}\\
&boxed{{color{#c34632} x=frac{35}{2} }}
end{align*}
$$

$$
{color{#4257b2}text{ e) }}
$$

$$
begin{align*}
&text{Subtract 3 from both sides}\
&frac{2}{3}x+3-3=frac{1}{4}x-7-3\\
&frac{2}{3}x=frac{1}{4}x-10 tag{Simplify}\
&frac{2}{3}x-frac{1}{4}x=frac{1}{4}x-10-frac{1}{4}xtag{Subtract 1/4x from both sides}\
&frac{5}{12}x=-10 tag{Simplify}\
&12cdot frac{5}{12}x=12left(-10right) tag{Multiply both sides by 12}\
&5x=-120 tag{Simplify}\
&frac{5x}{5}=frac{-120}{5} tag{Divide both sides by 5}\\
&boxed{{color{#c34632} x=-24 }}
end{align*}
$$

$$
{color{#4257b2}text{ f) }}
$$

$$
begin{align*}
&text{Multiply both sides by 3}\
&xcdot :3+frac{x}{3}cdot :3=4cdot :3\\
& 4x=12tag{Simplify}\
&frac{4x}{4}=frac{12}{4}tag{Divide both sides by 4}\\
&boxed{{color{#c34632} x=3 }}
end{align*}
$$

$$
color{#4257b2} text{a)} x= frac{23}{2}
$$

$$
color{#4257b2} text{b)No solution}
$$

$$
color{#4257b2} text{c)} 16
$$

$$
color{#4257b2} text{d)} x=frac{35}{2}
$$

$$
color{#4257b2} text{e)}x=-24
$$

$$
color{#4257b2} text{f)}x=3
$$

$textbf{(a)}$

$$
begin{align*}
t&=an+b tag{subtract $an$ from the each side}\
t-an&=an-an+b\
t-an&=btag{interchange sides}\
b&=t-an\
&boxed{b=t-an}
end{align*}
$$

$textbf{(b)}$

$$
begin{align*}
dfrac{y}{3}-a&=btag{ add $a$ to the each side}\
dfrac{y}{3}-a+a&=b+a\
dfrac{y}{3}&=b+atag{multiply each sideby 3}\
dfrac{y}{3}cdot 3&=(b+a)cdot 3\
y&=3(b+a)\
&boxed{y=3(b+a)}
end{align*}
$$

$textbf{(c)}$

$$
begin{align*}
F&=matag{ divide each side by $ a$}\
dfrac{F}{a}&=dfrac{ma}{a}\
dfrac{F}{a}&=m\
&boxed{m=dfrac{F}{a}}
end{align*}
$$

$textbf{(d)}$

$$
begin{align*}
F&=matag{ divide each side by $ m$}\
dfrac{F}{m}&=dfrac{ma}{m}\
dfrac{F}{m}&=a\
&boxed{a=dfrac{F}{m}}
end{align*}
$$

$textbf{(a)}$ $b=t-an$

$textbf{(b)}$ $y=3(b+a)$

$textbf{(c)}$ $m=dfrac{F}{a}$

$textbf{(d)}$ $a=dfrac{F}{m}$

$textbf{(a)}$ Estimation of the point of intersection of the line from the graph gives $(x,y)=(1.3 , 2.3)$

$textbf{(b)}$ First let’s consider the two points on the line with negative slope.
The two points on this line are $(0,6)$ and $(2,1)$

$$
text{Slope of this line is } =dfrac{1-6}{2-0}=-dfrac{5}{2}
$$

Now, equation of line with slope $-dfrac{5}{2}$ and passing through point $(0,6)$ is

$$
begin{align*}
y-6&=-dfrac{5}{2}(x-0)\
y&=-dfrac{5}{2}x+6
end{align*}
$$

$bullet$ Similarly, the two points on the line with positive slope are $(-1,0)$ and $(0,1)$

$$
text{Slope of this line is } =dfrac{1-0}{0-(-1)}=1
$$

Now, equation of line with slope $1$ and passing through point $(0,1)$ is

$$
begin{align*}
y-1&=1(x-0)\
y&=x+1
end{align*}
$$

$textbf{(c)}$ Solving for the point of intersection for given lines, we have

$$
begin{align}
y&=-dfrac{5}{2}x+6\
y&=x+1
end{align}
$$

From equation (2) we have $y=x+1$, hence putting this into the equation (1), we get

$$
begin{align*}
x+1&=-dfrac{5}{2}x+6\
x+dfrac{5}{2}x&=6-1\
dfrac{7}{2}x&=5\
x&=5times dfrac{2}{7}\
x&=dfrac{10}{7}\
x&=1.43\
Rightarrow y&=1.43+1\
y&=2.43
end{align*}
$$

We can see that the points via solving the equation and via estimations is very close and differ by 0.1 units.

$textbf{(a)}$ $(x,y)=(1.3 , 2.3)$

$textbf{(b)}$ $y=-dfrac{5}{2}x+6$ and $y=x+1$

$textbf{(c)}$ $(x,y)=(1.43,2.43)$

Given that the total number of members is 18 and total money to spent is $180.\
textbf{(a)}
Let’s assume number of t-shirts$=x$, and number of sweatshirts$=y$\$Rightarrow
$$
We have equations begin{align}
x+y&=18\
9x+15y&=180
end{align}\
textbf{(b)} Using the equation (1) , we can write
$$
x=18-y$

putting this x, into the equation (2), we get

$$
begin{align*}
9(18-y)+15y&=180\
162-9y+15y&=180\
6y&=180-162\
y&=dfrac{18}{6}\
y&=3\\
Rightarrow
x&=18-3\
x&=15
end{align*}
$$

Number of t-shirts $=15$

Number of sweatshirts $= 3$

$textbf{(a)}$

$$
begin{align*}
t(1)&=6\
t(2)&=18=6times 3\
t(3)&=54=18times 3
end{align*}
$$

We can see that each next term of the sequence can be obtained by multiplying the previous term by 3. Therefore, the multiplier for this sequence is 3.

$textbf{(b)}$

$$
begin{align*}
t(1)&=6=6cdot (3)^{1-1}\
t(2)&=18=6cdot 3=6cdot (3)^{2-1}\
t(3)&=54=6cdot 3 cdot 3=6cdot (3)^{3-1}\
t(n)&=6cdot (3)^{n-1}
end{align*}
$$

The explicit equation for this sequence is $boxed{t(n)=6cdot (3)^{n-1}}$.

$textbf{(c)}$

$$
begin{align*}
t(1)&=6\
t(2)&=18=3cdot 6=3cdot t(1)\
t(3)&=54=3cdot 18=3cdot t(2)\
t(n)&=3cdot t(n-1)
end{align*}
$$

The recursive equation for this sequence is $t(n)=3cdot t(n-1)$

$textbf{(a)}$ The multiplier for this sequence is 3.

$textbf{(b)}$ $t(n)=6cdot (3)^{n-1}$

$textbf{(c)}$ $t(n)=3cdot t(n-1)$

$textbf{(a)}$ Given average cost of the car is $31000 and price increase is 2% per year\$Rightarrow$price after one year$= 31000 +31000times $dfrac{2}{100}$=31000times(1.02)
$$
\
Hence, the multiplier is 1.02\\
textbf{(b)} with the continued trend the cost in 4 years will be
begin{align*}
text{cost in 1 year}&= 31000times 1.02\
text{cost in 2 year}&= (31000times 1.02)times 1.02\
text{cost in 3 year}&= ((31000times 1.02)times 1.02)times 1.02\
Rightarrow text{cost in n year}&= 31000times (1.02)^n tag{text{on generalising}}\
text{cost in 4 year}&= 31000times (1.02)^4\
text{cost in 4 year}&= $33555.4
end{align*}\
textbf{(c)} As derived above, cost in n year
$$
= $31000times (1.02)^n$
where, $1.02$ represents the multiplier per year including the 2% increase, and n represents the number of years.

$textbf{(a)}$ 1.02

$textbf{(b)}$ $33555.4\
textbf{(c)}$ $31000times (1.02)^n$