Problem

Solve for z 5/4(20z+12)=-35

The problem provided is a linear equation where you are required to find the value of the variable z. You need to apply the principles of algebra to isolate z on one side of the equation. This will typically involve distributing the fraction across the terms inside the parentheses, moving all terms with z to one side and constants to the other, and then dividing by the coefficient of z to solve for its value.

$\frac{5}{4} \left(\right. 20 z + 12 \left.\right) = - 35$

Answer

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Solution:

Step:1

Apply the inverse operation to both sides to isolate the term with $z$.

$$\frac{5}{4} \cdot \left(20z + 12\right) = -35 \times \frac{5}{4}$$

Step:2

Perform simplification on both sides.

Step:2.1

Focus on simplifying the left-hand side.

Step:2.1.1

Distribute the multiplication over addition.

$$\frac{5}{4} \cdot 20z + \frac{5}{4} \cdot 12 = -35 \times \frac{5}{4}$$

Step:2.1.2

Eliminate the common factors.

Step:2.1.2.1

Extract the factor of $4$ from $20z$.

$$\frac{5}{4} \cdot 4 \cdot 5z + \frac{5}{4} \cdot 12 = -35 \times \frac{5}{4}$$

Step:2.1.2.2

Reduce the fraction by cancelling out the $4$s.

$$5 \cdot 5z + \frac{5}{4} \cdot 12 = -35 \times \frac{5}{4}$$

Step:2.1.3

Carry out the multiplication of $5$ by $5z$.

$$25z + \frac{5}{4} \cdot 12 = -35 \times \frac{5}{4}$$

Step:2.1.4

Again, eliminate the common factors.

Step:2.1.4.1

Extract the factor of $4$ from $12$.

$$25z + \frac{5}{4} \cdot 4 \cdot 3 = -35 \times \frac{5}{4}$$

Step:2.1.4.2

Reduce the fraction by cancelling out the $4$s.

$$25z + 5 \cdot 3 = -35 \times \frac{5}{4}$$

Step:2.1.5

Complete the multiplication of $5$ by $3$.

$$25z + 15 = -35 \times \frac{5}{4}$$

Step:2.1.6

Apply the distributive property to the left-hand side.

$$25z + \frac{4}{5} \cdot 15 = -35 \times \frac{5}{4}$$

Step:2.1.7

Cancel out the common factor of $5$.

Step:2.1.7.1

Factor out $5$ from $25z$.

$$5 \cdot 5z + \frac{4}{5} \cdot 15 = -35 \times \frac{5}{4}$$

Step:2.1.7.2

Reduce the fraction by cancelling out the $5$s.

$$4 \cdot 5z + \frac{4}{5} \cdot 15 = -35 \times \frac{5}{4}$$

Step:2.1.8

Multiply $5$ by $4$.

$$20z + \frac{4}{5} \cdot 15 = -35 \times \frac{5}{4}$$

Step:2.1.9

Remove the common factor of $5$.

Step:2.1.9.1

Factor out $5$ from $15$.

$$20z + \frac{4}{5} \cdot 5 \cdot 3 = -35 \times \frac{5}{4}$$

Step:2.1.9.2

Reduce the fraction by cancelling out the $5$s.

$$20z + 4 \cdot 3 = -35 \times \frac{5}{4}$$

Step:2.1.10

Complete the multiplication of $4$ by $3$.

$$20z + 12 = -35 \times \frac{5}{4}$$

Step:2.2

Now, simplify the right-hand side.

Step:2.2.1

Simplify the multiplication.

Step:2.2.1.1

Cancel out the common factor of $5$.

$$20z + 12 = \frac{4}{5} \cdot 5 \cdot -7$$

Step:2.2.1.2

Reduce the fraction by cancelling out the $5$s.

$$20z + 12 = 4 \cdot -7$$

Step:2.2.1.3

Carry out the multiplication of $4$ by $-7$.

$$20z + 12 = -28$$

Step:3

Isolate the variable $z$ by moving all other terms to the opposite side.

Step:3.1

Subtract $12$ from both sides to remove the constant term from the left.

$$20z = -28 - 12$$

Step:3.2

Combine the constants on the right-hand side.

$$20z = -40$$

Step:4

Solve for $z$ by dividing both sides by the coefficient of $z$.

Step:4.1

Divide each term by $20$ to isolate $z$.

$$\frac{20z}{20} = \frac{-40}{20}$$

Step:4.2

Simplify the left-hand side by cancelling the $20$s.

$$z = \frac{-40}{20}$$

Step:4.3

Simplify the right-hand side by performing the division.

$$z = -2$$

Knowledge Notes:

To solve a linear equation, we often follow these steps:

  1. Distribute any multiplication over addition or subtraction to simplify the equation.

  2. Combine like terms on each side of the equation if necessary.

  3. Isolate the variable term on one side of the equation by adding or subtracting terms from both sides.

  4. Divide both sides by the coefficient of the variable to solve for the variable.

In this specific problem, we also used the property of inverses to cancel out the fraction on the left side by multiplying both sides by its reciprocal. Additionally, we used the distributive property to simplify expressions and cancelled common factors to reduce fractions.

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