Maths 1 Mark Question

m – 5 = 4
∴ m – 5 + 5 = 4 + 5
…. (Adding 5 to both sides)
∴ m + 0 = 9
∴ m = 9

x + 7 = 4
∴ x + 7 – 7 = 4 – 7 ….(Subtracting 7 from both sides)
∴ x + 0 = -3
∴ x = -3

\begin{array}{ll}
5(x+1)=74 \\
\therefore 5 x+5=74 \\
\therefore 5 x=74-5 \\
\therefore 5 x=69 \\
\therefore x =\frac{ 69 }{ 5 }
\end{array}

\begin{array}{ll}
2(x-4)=4 x+2 \\
\therefore 2 x-8=4 x+2 \\
\therefore -8-2=4 x-2 x \\
\therefore -10=2 x \\
\therefore x=\frac{-10}{2} \\
\therefore x=-5
\end{array}

\begin{aligned}
& 6 x -1=3 x +8 \\
& \therefore \quad 6 x-3 x=8+1 \\
& \therefore \quad 3 x=9 \\
& \therefore \quad x=\frac{9}{3} \\
& \therefore \quad x=3
\end{aligned}

\begin{array}{ll}
& 5 m -4=1 \\
\therefore & 5 m =1+4 \\
\therefore & 5 m =5 \\
\therefore & m =\frac{5}{5} \\
\therefore & m = 1
\end{array}

\begin{aligned}
& 10=2 y+5 \\
& \therefore \quad 10-5=2 y \\
& \therefore \quad 5=2 y \\
& \therefore \quad \frac{5}{2}=y \\
& \therefore \quad y=\frac{5}{ 2 }
\end{aligned}

\begin{array}{ll}
& 4 x+\frac{1}{2}=\frac{9}{2} \\
\therefore & 4 x=\frac{9}{2}-\frac{1}{2} \\
\therefore & 4 x=\frac{8}{2} \\
\therefore & 4 x=4 \\
\therefore & x=\frac{4}{4} \\
\therefore & x=1
\end{array}

4p = 12
\therefore \frac{4 p}{4}=\frac{12}{4} \ldots .(\text { Dividing both sides by } 4)
∴ p = 3

(-3y² + 10y – 16) + (7y² + 8)
= (-3y² + 7y²) + (10y) + (-16 + 8)
= 4y² + 10y – 8

(3y² – 10y + 16) + (2y – 7)
= 3y² + (-10y + 2y) + (16 – 7)
= 3y² – 8y + 9

(17a²b² + 16c) + (28c – 28a²b²)
= [17a²b² + (-28a²b²)] + (16c + 28c)
= -11a²b² + 44c

(13x² – 12y²) + (6x² – 8y²)
= (13x² + 6x²) + [(-12y²) + (-8y²)]
= 19x² + (-20y²)
= 19x² – 20y²

(2a + 6b + 8c) + (16a + 13c + 18b)
= (2a + 16a) + (6b + 18b) + (8c + 13c)
= 18a + 24b + 21c

(9p + 16q) + (13p + 2q)
= (9p + 13p) + (16q + 2q)
= 22p + 18q