Addition, Subtraction, Multiplication and Division: Using variables

We begin by placing parentheses around the two multiplication exercises:

$(20\times y)+(8\times2)-7=14$

We then solve the exercises within the parentheses:

$20y+16-7=14$

We simplify:

$20y+9=14$

We move the sections:

$20y=14-9$

$20y=5$

We divide by 20:

$y=\frac{5}{20}$

$y=\frac{5}{5\times4}$

We simplify:

$y=\frac{1}{4}$

To solve this problem, we'll follow these steps:

• Simplify the term $2y \cdot \frac{1}{y}$
• Rearrange the equation to group similar terms
• Solve for $y$

Now, let's work through each step:

Step 1: Simplify the expression $2y \cdot \frac{1}{y}$.

The term $2y \cdot \frac{1}{y}$ simplifies directly to $2$ since $y$ in the numerator and denominator cancel each other out assuming $y \neq 0$. Therefore, the equation becomes:

$2 - y + 4 = 8y$

Step 2: Combine like terms on the left-hand side:

$2 + 4 = 6$, so the equation now is $6 - y = 8y$.

Step 3: Rearrange the equation to isolate $y$ on one side. Add $y$ to both sides to get rid of the negative $y$:

$6 = 8y + y$

This simplifies to:

$6 = 9y$

Step 4: Solve for $y$ by dividing both sides by 9:

$y = \frac{6}{9}$

Simplify the fraction to get:

$y = \frac{2}{3}$

Therefore, the solution to the problem is $\frac{2}{3}$.

We begin by placing the multiplication exercise inside of parentheses:

$-25-42\colon x+(18\times2)\colon4=-23$

$-25-42\colon x+36\colon4=-23$

We will then place the division exercise inside of parentheses:

$-25-42\colon x+(36\colon4)=-23$

$-25-42\colon x+9=-23$

Next we rearrange the exercise in order to simplify it:

$-25+9-42\colon x=-23$

$(-25+9)-42\colon x=-23$

We then solve the exercise inside of the parenthesis and obtain the following:

$-16-42\colon x=-23$

We rearrange the fractions and obtain the following:

$-42\colon x=-23+16$

$-42\colon x=-7$

We multiply by x and obtain the following:

$-42=-7x$

Lastly we divide by negative 7:

$x=\frac{-42}{-7}=6$