Solving First-Degree Equations Practice Problems - All Methods

To solve the equation $5 - x = 4$, we aim to isolate $x$ on one side of the equation.

We start by considering the equation:
$5 - x = 4$

Step 1: Eliminate 5 from the left side to isolate terms involving $x$. To do this, subtract 5 from both sides of the equation:

$(5 - x) - 5 = 4 - 5$

Step 2: Simplify both sides:

$-x = -1$

Step 3: To solve for $x$, multiply or divide both sides by $-1$ to change the sign of $x$:

$-1 \cdot -x = -1 \cdot -1$

This simplifies to:

$x = 1$

Therefore, the solution to the equation $5 - x = 4$ is $x = 1$.

The correct answer is $x = 1$.

To solve for $b$, we need to isolate it on one side of the equation. Starting with:

$b+6=14$

Subtract$6$ from both sides to get:

$b+6-6=14-6$

This simplifies to:

$b=8$

Therefore, the solution is $b = 8$.

We use the formula:

$a\cdot x=b$

$x=\frac{b}{a}$

We multiply the numerator by X and write the exercise as follows:

$\frac{x}{4}=3$

We multiply by 4 to get rid of the fraction's denominator:

$4\times\frac{x}{4}=3\times4$

Then, we remove the common factor from the left side and perform the multiplication on right side to obtain:

$x=12$

We open the parentheses according to the formula:

$a(x+b)=ax+ab$

$5\times x+5\times3=0$

$5x+15=0$

We will move the 15 to the right section and keep the corresponding sign:

$5x=-15$

Divide both sections by 5

$\frac{5x}{5}=\frac{-15}{5}$

$x=-3$

To open parentheses we will use the formula:

$a(x+b)=ax+ab$

$(7\times-2x)+(7\times5)=77$

We multiply accordingly

$-14x+35=77$

We will move the 35 to the right section and change the sign accordingly:

$-14x=77-35$

We solve the subtraction exercise on the right side and we will obtain:

$-14x=42$

We divide both sections by -14

$\frac{-14x}{-14}=\frac{42}{-14}$

$x=-3$