Question

1. solve the following equations.

i. $12 = 5 - 2c$
a. $c = 8.5$
b. $c = -3.5$
c. $c = 3.5$
d. $c = -8.5$
ii. $-4 = \frac{x + 3}{5}$
a. $x = -17$
b. $x = 17$
c. $x = 23$
d. $x = -23$
iii. $5x^3 + 6 = 46$
a. $x = 2$
b. $x = -2$
c. $x = \pm 2$
d. $x = 3$

1. solve the following equations.

i. $8x^3 - 6x^3 + 4 = 132$
a. $x = -4$
b. $x = \pm 4$
c. $x = 3$
d. $x = 4$
ii. $4(2x - 8) - 3(4 - 5x) = 5$
a. $x = \frac{49}{23}$
b. $x = \frac{23}{49}$
c. $x = -\frac{49}{23}$
d. $x = -\frac{23}{49}$

Explanation:

Problem 2. I (Solving \( 12 = 5 - 2c \))

Step 1: Isolate the term with \( c \)

Subtract 5 from both sides: \( 12 - 5 = -2c \)
\( 7 = -2c \)

Step 2: Solve for \( c \)

Divide both sides by -2: \( c=\frac{7}{-2}=-3.5 \)

Step 1: Eliminate the denominator

Multiply both sides by 5: \( -4\times5=x + 3 \)
\( -20=x + 3 \)

Step 2: Solve for \( x \)

Subtract 3 from both sides: \( x=-20 - 3=-23 \)

Step 1: Isolate the cubic term

Subtract 6 from both sides: \( 5x^{3}=46 - 6 \)
\( 5x^{3}=40 \)

Step 2: Solve for \( x^{3} \)

Divide both sides by 5: \( x^{3}=\frac{40}{5}=8 \)

Step 3: Solve for \( x \)

Take the cube root of both sides: \( x=\sqrt[3]{8}=2 \)

Answer:

b. \( c = -3.5 \)

Problem 2. II (Solving \( -4=\frac{x + 3}{5} \))