Question

name: date: period:
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. $6x + 1 - 4x = x + 7$; $x = 8$
3. $3(x - 4) + 2 = 2x - 6$; $x = 2$
4. $7x - 5 = 2(3x + 1)$; $x = -3$

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 and substitute \( x = 8 \).

First, simplify the left - hand side of the equation \( 6x + 1-4x=(6x-4x)+1 = 2x+1 \). Then substitute \( x = 8 \): \( 2\times8+1=16 + 1=17 \)

Step 2: Substitute \( x = 8 \) into the right - hand side (RHS) of the equation.

The right - hand side of the equation is \( x + 7 \). Substitute \( x = 8 \): \( 8+7 = 15 \)

Step 3: Compare LHS and RHS.

Since \( 17
eq15 \) (LHS = 17 and RHS = 15), LHS \(
eq\) RHS.

Step 1: Substitute \( x = 2 \) into the left - hand side (LHS) of the equation.

First, expand the left - hand side: \( 3(x - 4)+2=3x-12 + 2=3x-10 \). Substitute \( x = 2 \): \( 3\times2-10=6-10=-4 \)

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

The right - hand side of the equation is \( 2x-6 \). Substitute \( x = 2 \): \( 2\times2-6=4-6=-2 \)

Step 3: Compare LHS and RHS.

Since \( - 4
eq-2 \) (LHS \(=-4\) and RHS \(=-2\)), LHS \(
eq\) RHS.

Answer:

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

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