Question

week 5 packet
part a: checking solutions by substitution
directions: substitute the given value of x into both sides of the equation. if both sides are equal, the value is a solution.

1. ( 5x - 3 = 4(2x + 3) ); ( x = -5 )
2. ( 6x + 1 - 4x = x + 7 ); ( x = 8 )

Explanation:

Problem 1: \( 5x - 3 = 4(2x + 3) \); \( x = -5 \)

Step 1: Substitute \( x = -5 \) into the left - hand side (LHS) of the equation.

The left - hand side of the equation is \( 5x-3 \). Substitute \( x=-5 \):
\( 5\times(-5)-3=-25 - 3=-28 \)

Step 2: Substitute \( x = -5 \) into the right - hand side (RHS) of the equation.

The right - hand side of the equation is \( 4(2x + 3) \). First, substitute \( x=-5 \) into the expression inside the parentheses: \( 2\times(-5)+3=-10 + 3=-7 \). Then multiply by 4: \( 4\times(-7)=-28 \)

Step 3: Compare LHS and RHS.

Since LHS \(=-28\) and RHS \(=-28\), LHS = RHS.

Step 1: Simplify the left - hand side (LHS) of the equation.

Combine like terms on the LHS: \( 6x-4x + 1=2x + 1 \)

Step 2: Substitute \( x = 8 \) into the LHS.

Substitute \( x = 8 \) into \( 2x+1 \): \( 2\times8+1=16 + 1=17 \)

Step 3: Substitute \( x = 8 \) into the right - hand side (RHS) of the equation.

The RHS of the equation is \( x + 7 \). Substitute \( x = 8 \): \( 8+7 = 15 \)

Step 4: Compare LHS and RHS.

Since LHS \(=17\) and RHS \(=15\), LHS \(
eq\) RHS.

Answer:

\( x = - 5 \) is a solution.

Problem 2: \( 6x + 1-4x=x + 7 \); \( x = 8 \)