Question 1

We will find the value of the unknown variable in each equation. Then we will check if our answer is correct.

Step 1 — Solving equation (a)

Let us solve the first equation. The equation is $3x - 10 = 35$. We want to get $x$ by itself. First, we add 10 to both sides of the equation. This will remove the -10 from the left side.

$3x - 10 + 10 = 35 + 10$

$3x = 45$

Now, we need to get $x$ alone. We divide both sides by 3. This will undo the multiplication by 3 on the left side.

$\frac{3x}{3} = \frac{45}{3}$

$\boxed{x = 15}$

Step 2 — Checking solution (a)

Let us check if $x = 15$ is correct. We put 15 in place of $x$ in the original equation. The left side of the equation is $3x - 10$.

$\text{LHS} = 3 \times 15 - 10$

$= 45 - 10$

$= 35$

The right side of the equation is 35. Since the left side equals the right side, our answer is correct.

$\text{LHS} = \text{RHS}$

Step 3 — Solving equation (b)

Let us solve the second equation. The equation is $5s = 3s$. We want to find the value of $s$. We need to bring all terms with $s$ to one side. We subtract 3s from both sides of the equation.

$5s - 3s = 3s - 3s$

$2s = 0$

Now, we divide both sides by 2. This will isolate $s$.

$\frac{2s}{2} = \frac{0}{2}$

$\boxed{s = 0}$

Step 4 — Checking solution (b)

Let us check if $s = 0$ is correct. We put 0 in place of $s$ in the original equation. The left side of the equation is $5s$.

$\text{LHS} = 5 \times 0$

$= 0$

The right side of the equation is $3s$.

$\text{RHS} = 3 \times 0$

$= 0$

Since the left side equals the right side, our answer is correct.

$\text{LHS} = \text{RHS}$

Step 5 — Solving equation (c)

Let us solve the third equation. The equation is $3u - 7 = 2u + 3$. We want to get $u$ by itself. First, we move the $u$ terms to one side. We subtract 2u from both sides.

$3u - 7 - 2u = 2u + 3 - 2u$

$u - 7 = 3$

Now, we move the constant terms to the other side. We add 7 to both sides.

$u - 7 + 7 = 3 + 7$

$\boxed{u = 10}$

Step 6 — Checking solution (c)

Let us check if $u = 10$ is correct. We put 10 in place of $u$ in the original equation. The left side of the equation is $3u - 7$.

$\text{LHS} = 3 \times 10 - 7$

$= 30 - 7$

$= 23$

The right side of the equation is $2u + 3$.

$\text{RHS} = 2 \times 10 + 3$

$= 20 + 3$

$= 23$

Since the left side equals the right side, our answer is correct.

$\text{LHS} = \text{RHS}$

Step 7 — Solving equation (d)

Let us solve the fourth equation. The equation is $4(m + 6) - 8 = 2m - 4$. First, we use the distributive property on the left side. We multiply 4 by both $m$ and 6.

$4m + 4 \times 6 - 8 = 2m - 4$

$4m + 24 - 8 = 2m - 4$

Now, we combine the constant numbers on the left side.

$4m + 16 = 2m - 4$

Next, we move the $m$ terms to one side. We subtract 2m from both sides.

$4m + 16 - 2m = 2m - 4 - 2m$

$2m + 16 = -4$

Now, we move the constant terms to the other side. We subtract 16 from both sides.

$2m + 16 - 16 = -4 - 16$

$2m = -20$

Finally, we divide both sides by 2. This will give us the value of $m$.

$\frac{2m}{2} = \frac{-20}{2}$

$\boxed{m = -10}$

Step 8 — Checking solution (d)

Let us check if $m = -10$ is correct. We put -10 in place of $m$ in the original equation. The left side of the equation is $4(m + 6) - 8$.

$\text{LHS} = 4(-10 + 6) - 8$

$= 4(-4) - 8$

$= -16 - 8$

$= -24$

The right side of the equation is $2m - 4$.

$\text{RHS} = 2(-10) - 4$

$= -20 - 4$

$= -24$

Since the left side equals the right side, our answer is correct.

$\text{LHS} = \text{RHS}$

Step 9 — Solving equation (e)

Let us solve the fifth equation. The equation is $\frac{u}{15} = 6$. We want to get $u$ by itself. The $u$ is being divided by 15. To undo this, we multiply both sides by 15.

$\frac{u}{15} \times 15 = 6 \times 15$

$\boxed{u = 90}$

Step 10 — Checking solution (e)

Let us check if $u = 90$ is correct. We put 90 in place of $u$ in the original equation. The left side of the equation is $\frac{u}{15}$.

$\text{LHS} = \frac{90}{15}$

$= 6$

The right side of the equation is 6. Since the left side equals the right side, our answer is correct.

$\text{LHS} = \text{RHS}$

Answer

(a) $x = 15$ (b) $s = 0$ (c) $u = 10$ (d) $m = -10$ (e) $u = 90$