$3x^2-7x+5+x^2-x+3$

$$
=4x^2-8x+8
$$

$x^2-6x+1+x^2+6x-5$

$$
=2x^2-4
$$

$2x^2-4x+3-x^2+3x-2+x^2-1$

$$
=2x^2
$$

(a) $4x^2-8x+8$

(b) $2x^2-4$

(c) $2x^2$

Simplify $f(x)$ and $g(x)$ by combining similar terms.

Remember that $a-(c-d+e)=a+(-c+d-e)$

$f(x)=(2x-1)-(3-4x)+(x+2)$

$f(x)=(2x-1)+(-3+4x)+(x+2)$

$f(x)=(2x+4x+x)+(-1-3+2)$

$f(x)=7x-2$

$g(x)=(-x+6)+(6x-9)-(-2x-1)$

$g(x)=(-x+6)+(6x-9)+(2x+1)$

$g(x)=(-x+6x+2x)+(6-9+1)$

$g(x)=7x-2$

Therefore, $f(x)=g(x)$

$f(x)=7x-2$

$g(x)=7x-2$

$$
therefore f(x)=g(x)
$$

You can choose any value of $x$ to show the non-equivalence except for $x=3$ because $f(3)=g(3)$.

We can choose $x=0$

$f(x)=2(x-3)+3(x-3)$

$f(0)=2(0-3)+3(0-3)$

$f(0)=2(-3)+3(-3)=-6-9=-15$

$g(x)=5(2x-6)$

$g(0)=5[2(0)-6]$

$g(0)=5(0-6)=-30$

Therefore, $f(x)neq g(x)$

$f(0)=-15$

$g(0)=-30$

$$
therefore f(x)neq g(x)
$$

Simplify each expression by combining similar terms.

a) $(2a+4c+8)+(7a-9c-3)$

$=(2a+7a)+(4c-9c)+(8-3)$

$=9a-5c+5$

b) $(3x+4y-5z)+(2x^2+6z)$

$=3x+4y+2x^2+(-5x+6z)$

$=3x+4y+2x^2+z$

c) $(6x+2y+9)+(-3x-5y-8)$

$=(6x-3x)+(2y-5y)+(9-8)$

$=3x-3y+1$

d) $(2x^2-7x+6)+(x^2-2x-9)$

$=(2x^2+x^2)+(-7x-2x)+(6-9)$

$=3x^2-9x-3$

e) $(-4x^2-2xy)+(6x^2-3xy+2y^2)$

$=(-4x^2+6x^2)+(-2xy-3xy)+2y^2$

$=2x^2-5xy+2y^2$

f) $(x^2+y^2+8)+(4x^2-2y^2-9)$

$=(x^2+4x^2)+(y^2-2y^2)+(8-9)$

$=5x^2-y^2-1$

a) $9a-5c+5$

b) $2x^2+3x+4y+z$

c) $3x-3y+1$

d) $3x^2-9x-3$

e) $2x^2-5xy+2y^2$

f) $5x^2-y^2-1$

Simplify the following expressions by combining similar terms. Also be careful in subtracting algebraic expressions.

$$
begin{equation*}a-(b-c+d)=a+(-b+c-d)end{equation*}
$$

$$
begin{align*}
bold{a)} ;; (m-n+2p)-(3n+p-7)
&= m-n+2p+(-3n-p+7)\
&= m+(-n-3n)+(2p-p)+7\
&= boxed{m-4n+p+7}
end{align*}
$$

$$
begin{align*} bold{b)};; (-6m-2q+8)-(2m+2q+7)
&=(-6m-2q+8)+(-2m-2q-7)\
&=(-6m-2m)+(-2q-2q)+(8-7)\
&=boxed{-8m-4q+1}
end{align*}
$$

$$
begin{align*} bold{c)};; (4a^2-9)-(a^3+2a-9)
&=(4a^2-9)+(-a^3-2a+9)\
&=4a^2-a^3-2a+(-9+9)\
&=boxed{4a^2-a^3-2a}
end{align*}
$$

$$
begin{align*} bold{d)};; (2m^2-6mn+8n^2)\-(4m^2-mn-7n^2)
&=(2m^2-6mn+8n^2)+(-4m^2+mn+7n^2)\
&=(2m^2-4m^2)+(-6mn+mn)+(8n^2+7n^2)\
&=boxed{-2m^2-5mn+15n^2}
end{align*}
$$

$$
begin{align*} bold{e)} ;;(3x^2+2y^2+7)-(4x^2-2y^2-8)
&=(3x^2+2y^2+7)+(-4x^2+2y^2+8)\
&=(3x^2-4x^2)+(2y^2+2y^2)+(7+8)\
&=boxed{-x^2+4y^2+15}\
end{align*}
$$

$$
begin{align*} bold{f)} ;;5x^2-(2x^2-30)-(-20)
&=5x^2+(-2x^2+30)+(20)\
&=(5x^2-2x^2)+(30+20)\
&=boxed{3x^2+50}
end{align*}
$$

a) $m-4n+p+7$

b) $-8m-4q+1$

c) $-a^3+4a^2-2a$

d) $-2m^2-5mn+15n^2$

e) $-x^2+4y^2+15$

f) $3x^2+50$

Simplify the following expressions by combining similar terms. Also be careful in subtracting algebraic expressions.

$$
begin{equation*}a-(b-c+d)=a+(-b+c-d)end{equation*}
$$

$$
begin{align*}
bold{a)};;;;(2x-y)-(-3x+4y)+(6x-2y)&= (2x-y)+(3x-4y)+(6x-2y)\
&=(2x+3x+6x)+(-y-4y-2y)\
&=boxed{11x-7y}\
end{align*}
$$

$$
begin{align*}
bold{b)};;;;(3x^2-2x)+(x^2-7x)-(7x+3)
&= (3x^2+x^2)+(-2x-7x-7x)+3\
&=boxed{4x^2-16x+3}\
end{align*}
$$

$$
begin{align*}
bold{c)};;;;&(2x^2+xy-y^2)-(x^2-4xy-y^2)+(3x^2-5xy) \
&;;= (2x^2+xy-y^2)+(-x^2+4xy+y^2)+(3x^2-5xy)\
&;;=(2x^2-x^2+3x^2)+(xy+4xy-5xy)+(-y^2+y^2)\
&;;=boxed{4x^2}\
end{align*}
$$

$$
begin{align*}
bold{d)};;;;&(xy-xz+4yz)+(2x-3yz)-(4y-xz)\
&;;;;= (xy-xz+4yz)+(2x-3yz)+(-4y+xz)\
&;;;;=xy+(-xz+xz)+(4yz-3yz)+2x-4y\
&;;;;=xy+0+yz+2x-4y\
&;;;;=boxed{xy+yz+2x-4y}\
end{align*}
$$

$$
begin{align*} bold{e)};;;;left( frac{1}{2}x+dfrac{1}{3}yright)-left(dfrac{1}{5}x-yright)
&= left(frac{1}{2}x+frac{1}{3}yright)+left(-frac{1}{5}x+yright)\
&= left(frac{1}{2}x-frac{1}{5}xright)+left(frac{1}{3}y+yright)\
&= left(frac{5}{10}x-frac{2}{10}xright)+left(frac{1}{3}y+frac{3}{3}yright)\
&= boxed{ dfrac{3}{10}x+dfrac{4}{3}y}\
end{align*}
$$

$$
begin{align*} bold{f)};;;;left(dfrac{3}{4}x+dfrac{1}{2}yright)-left(dfrac{2}{3}x+dfrac{1}{4}y-1right)
&= left(dfrac{3}{4}x+dfrac{1}{2}yright)+left(-dfrac{2}{3}x-dfrac{1}{4}y+1right)\
&= left(dfrac{3}{4}x-dfrac{2}{3}xright)+left(dfrac{1}{2}y-dfrac{1}{4}yright)+1\
&= left(dfrac{3(3)}{4(3)}x-dfrac{2(4)}{3(4)}xright)+left(dfrac{1(2)}{2(2)}y-dfrac{1}{4}yright)+1\
&= left(dfrac{9}{12}x-dfrac{8}{12}xright)+left(dfrac{2}{4}y-dfrac{1}{4}yright)+1\
&=boxed{dfrac{1}{12}x+dfrac{1}{4}y+1}\
end{align*}
$$

a) $11x-5y$

b) $4x^2-16x-3$

c) $4x^2$

d) $xy+yz+2x-4y$

e) $dfrac{3}{10}x+dfrac{4}{3}y$

f) $dfrac{1}{12}x+dfrac{1}{4}y+1$

$3x^2-x-5x^2+xneq-2x^2-2x$

$$
-2x^2neq-2x^2-2x
$$

$3(1)^2-1-5(1)^2+1neq-2(1)^2-2(1)$

$3-1-5+1neq-2-2$

$$
-3neq-4
$$

Let $x=1$ and plug into the equations setting them equal to one another

$$
begin{align*}
bold{a)};;;; f(x) &= (2x^2+7x-2)-(3x+7)\
&= (2x^2+7x-2)+(-3x-7)\
&=(2x^2)+(7x-3x)+(-2-7)\
&=2x^2+4x-9\\
g(x) &= (x^2+12)+(x^2+4x-17)\
&=(x^2+x^2)+4x+(12-17)\
&=2x^2+4x-5\\
&text{Therefore}, f(x)neq g(x)
end{align*}
$$

$bold{b)}$ Remember the special product $(x+y)^3=x^3+3x^2y+3xy^2+y^3$

$$
begin{align*}
s_1(t) &= (t+2)^3\
&=t^3+3t^2(2)+3(t)(2^2)+2^3\
&=t^3+6t^2+12t+8\
&neq t^3+8\\
&text{Therefore}, s_1(t)neq s_2(t)
end{align*}
$$

$bold{c)}$ Remember the distributive property $a(b+c)=ab+ac$

$$
begin{align*} y_1 &= (x-1)(x)(x+2)\
&= (x-1)(x^2+2x)\
&=x^2(x-1)+2x(x-1)\
&=x^3-x^2+2x^2-2x\
&=x^3+x^2-2x\\
y_2&=3x(x^2-1)\
&=3x(x^2)-3x(1)\
&=3x^3-3x\\
&text{Therefore}, y_1neq y_2\
end{align*}
$$

$bold{d)}$ Be careful in subtracting expressions $a-(b+c-d)=a+(-b-c+d)$

$$
begin{align*} f(n) &= 0.5n^2+2n-3+(1.5n^2-6)\
&= (0.5n^2+1.5n^2)+2n+(-3-6)\
&=2n^2+2n-9\\
g(n) &= n^2-n+1-(-n^2-3n+10)\
&= n^2-n+1+(n^2+3n-10)\
&= n^2-n+1+n^2+3n-10\
&= (n^2+n^2)+(-n+3n)+(1-10)\
&=2n^2+2n-9\\
&text{Therefore}, f(n)=g(n)\
end{align*}
$$

$bold{e)}$ Use distributive property $a(b+c)=ab+ac$

$$
begin{align*} y_1&=3p(q-2)+2p(q+5)\
&= (3p)(q)+(3p)(-2)+(2p)q+(2p)(5)\
&= 3pq-6p+2pq+10p\
&= (3pq+2pq)+(-6p+10p)\
&= 5pq+4p\\
y_2&=p(q+4)\
&= pq+4p\\
&text{Therefore}, y_1 neq y_2\
end{align*}
$$

$bold{f)}$ Use distributive property $a(b+c)=ab+ac$

$$
begin{align*} f(m) &= m(5-m)-2(2m-m^2)\
&= 5m-m^2-(4m-2m^2)\
&= 5m-m^2+(-4m+2m^2)\
&= 5m-m^2-4m+2m^2\
&= (5m-4m)+(-m^2+2m^2)\
&= m+m^2\\
g(m) &= 4m^2(m-1)-3m^2+5m\
&= 4m^3-4m^2-3m^2+5m\
&= 4m^3+(-4m^2-3m^2)+5m\
&= 4m^3-7m^2+5m\\
&text{Therefore, } f(m)ne g(m)
end{align*}
$$

a) $f(x)neq g(x)$

b) $s_1(t)neq s_2(t)$

c) $y_1neq y_2$

d) $f(n)=g(n)$

e) $y_1neq y_2$

f) $f(m)neq g(m)$

We must find non-equivalent polynomial such that $f(0)=g(0)$ and $f(2)=g(2)$

This is easier to do in factored form such that $f(0)=0$ and $f(2)=0$ implying that $(x)(x-2)$ must be a factor. To make them non-equivalent, we must multiply it when another factor that is different for $f(x)$ and $g(x)$

Here are some examples:

$f(x)=x(x-2)(x+1)$

$g(x)=x(x-2)(x+3)$

$f(x)=x(x-2)(3x+5)$

$g(x)=x(x-2)(5x+3)$

$f(x)=x(x-2)(x^2-2)$

$g(x)=x(x-2)(x^3+2x)$

$f(x)=x(x-2)(x+1)$

$g(x)=x(x-2)(x+3)$

With $x$ correct answers from a total of $25$ items, $y$ questions unanswered, $6$ points for a correct answer, $2$ points for not answering a question, and $1$ point for an incorrect answer, then

a) Kosuke’s number of questions answered incorrectly is

$$
begin{align*}
(25-x-y)text{ items}
.end{align*}
$$

b) Kosukes score is

$$
begin{align*}
&
6(x)+2(y)+1(25-x-y)
\&=
6x+2y+25-x-y
\&=
(5x+y+25)text{ points}
.end{align*}
$$

c) Using the result of (b), then with $x=13$ and $y=25-13-7=5,$ Kosuke’s score is

$$
begin{align*}
&
5x+y+25
\&=
5(13)+5+25
\&=
65+5+25
\&=
95text{ points}
.end{align*}
$$

a) $(25-x-y)$ items

b) $(5x+y+25)$ points

c) $95$ points

$2x+3y-1+2x+3y-1+w=7x+9y$

$4x+6y-2+w=7x+9y$

$$
w=3x+3y+2
$$

The perimeter is the sum of the side lengths. Let the unknown side lengths for which we are driving to solve be $w$. Combine like terms and simplify

$$
3x+3y+2
$$

a) If $R(x)$ is revenue function and $C(x)$ is the cost function, then the profit function is $P(x)=R(x)-C(x)$

$$
begin{align*}
P(x) &= R(x)-C(x)\
&= (-50x^2+2500x)-(150x+9500)\
&= (-50x^2+2500x)+(-150x-9500)\
&= -50x^2+(2500x-150x)-9500\
P(x) &=-50x^2+2350x-9500\
end{align*}
$$

b) We shall find the profit $P(x)$ when $x=12$

$$
begin{align*} P(12)&=-50(12^2)+2350(12)-9500\
&=-50(144)+2350(12)-9500\
P(12)&=$11500 end{align*}
$$

a) $P(x)=-50x^2+2350x-9500$

b) $P(12)=$11500$

Two functions are said to be equivalent if they have the same value for all values of $x$ in their domain. If they have different value for at least one value of $x$, then they are not equivalent. If they are equal only for some values of $x$, then it cannot be determined.

$bold{a);;}$ The functions are equal to some value of $x$ but we have no information whether they are equal for other values of $x$, thus, it cannot be determined.

$bold{b);;}$ We cannot conclude that they are non-equivalent because it is possible that $h(3)=h(4)$ and $g(3)=g(4)$ which lead to $h(3)=g(4)$. Thus, it cannot be determined.

$bold{c);;}$ At the same value of $x$, the two functions have different values, so it must be non-equivalent.

$bold{d);;}$ At the same value of $x$, the two functions have different values, so it must be non-equivalent.

$bold{e);;}$ $n(x)=p(x)$ for all $x$, so it must be equivalent.

$bold{a)}$ Here we need to prove that the pattern is true for any number on the calendar. Since there are 7 days in a week, each column is made of numbers whose first differences is 7. \\
Let $x$ be any number on the calendar.\\
begin{tabular}{|l|ll}
cline{1-1}
$x$ & & \ cline{1-1}
$x+7$ & & \ hline
multicolumn{1}{|c|}{$x+14$} & multicolumn{1}{c|}{$x+15$} & multicolumn{1}{c|}{$x+16$} \ hline
end{tabular}\\\
vertical entries (column): $x$, $x+7$, $x+14$\\
corner number: $x+14$\\
horizontal entries (row): $x+14$, $x+15$, $x+16$\\
The patterns says the sum of the five numbers is equal to five times the corner number minus 18.\\
sum of five numbers: begin{equation*}(x)+(x+7)+(x+14)+(x+15)+(x+16)=boxed{5x+52}end{equation*}
five times the corner number minus: begin{equation*} 18 = 5(x+14)-18=5x+70-18=boxed{5x+52}end{equation*}\
Since both expressions are equal to $5x+52$, this is true for any number $x$ on the calendar.\\

$bold{b);;}$ We have shown in part(a) that the sum of the numbers in this pattern is $5x+52$ and the corner number is $x+14$, here we need to find value of $x+14$ such that $5x+52=112$. To do that, we must find the value of $x$ first.

$$
begin{align*} 5x+52&=112\
5x&=112-52 \
5x&=60\
x&=12 end{align*}
$$

The corner number is $x+14=12+14=boxed{26}$

$bold{c);;}$ If $x$ is the corner number instead of $x+14$, the sum of the five numbers is obtained by replacing $x$ with $x-14$ \\
$5(x-14)+52=5x-70+52=boxed{5x-18}$\\
Alternatively, we can predict from the pattern that the entries would be:\\
begin{tabular}{|l|ll}
cline{1-1}
$x-14$ & & \ cline{1-1}
$x-7$ & & \ hline
multicolumn{1}{|c|}{$x$} & multicolumn{1}{c|}{$x+1$} & multicolumn{1}{c|}{$x+2$} \ hline
end{tabular}\\
So the sum is\\
$x+(x+1)+(x+2)+(x-7)+(x-14)=boxed{5x-18}$\\

a) $x+(x+7)+(x+14)+(x+15)+(x+16)=5(x+14)-18$

b) 26

c) $5x-18$

$bold{a);;}$ We need to find 5 consecutive natural numbers whose sum is $105$. Consecutive natural numbers is obtained by adding 1 to the previous number. If $x$ is the first number,

$$
begin{align*} x+(x+1)+(x+2)+(x+3)+(x+4)&=105\
5x+10&=105\
5x&=95\
x&=19
end{align*}
$$

Therefore, the give consecutive numbers are

$$
19+20+21+22+23=105
$$

$bold{ b);;}$ Notice from part(a) that the middle term $21$ is $dfrac{105}{5}$

This suggests that if $m=dfrac{n}{5}$, then $m$ must be the middle term which leads to

$$
n=(m-2)+(m-1)+(m)+(m+1)+(m+2)
$$

$bold{c);;}$ We will do similar procedure as in part(a)

$$
begin{align*}
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)&=91\
7x+21&=91\
7x&=70\
x&=10
end{align*}
$$

Therefore, $10+11+12+13+14+15+16=91$

a) $19+20+21+22+23$

b) $n=(m-2)+(m-1)+m+(m+1)+(m+2)$

c) $10+11+12+13+14+15+16$

Substituting $x=2$ in the given functions, then

$$
begin{align*}
f(x)&=ax+b
\
f(2)&=a(2)+b
\
f(2)&=2a+b
\\text{and}\\
g(x)&=cx+d
\
g(2)&=c(2)+d
\
g(2)&=2c+d
.end{align*}
$$

Since it is given that $f(2)=g(2),$ then

$$
begin{align}
2a+b=2c+d
.end{align}
$$

Substituting $x=5$ in the given functions, then

$$
begin{align*}
f(x)&=ax+b
\
f(5)&=a(5)+b
\
f(5)&=5a+b
\\text{and}\\
g(x)&=cx+d
\
g(5)&=c(5)+d
\
g(5)&=5c+d
.end{align*}
$$

Since it is given that $f(5)=g(5),$ then

$$
begin{align}
5a+b=5c+d
.end{align}
$$

Subtracting equations (1) and (2) results to

$$
begin{align*}
2a+b-(5a+b)&=2c+d-(5c+d)
\
2a+b-5a-b&=2c+d-5c-d
\
-3a&=-3c
\
a&=c
.end{align*}
$$

Using $a=c$ in equation (1) results to

$$
begin{align*}
2a+b&=2c+d
\
2c+b&=2c+d
\
2c-2c+b&=d
\
b&=d
.end{align*}
$$

Since $a=c$ and $b=d,$ then

$$
begin{align*}
f(x)&=ax+b
\&=
cx+d
\&=g(x)
.end{align*}
$$

Hence, $f(x)$ and $g(x)$ are equivalent.