Question

1. for each equation, determine whether it has one solution, no solution, or infinitely many solutions by placing a check mark in the box.

equation \tone solution \tno solution \tinfinitely many solutions
$x + 2 = x + 4$
$2x + 4 = 2x + 1 + 3$
$3(x - 1) = 2x + 1$
$-(4x - 5) = -x + 5 - 3x$

1. if an equation is never true for any value of $x$, which is true about the equation?

a. it has no solution.
b. it has one solution.
c. it has infinitely many solutions.
d. zero is its only solution.

1. for each equation, decide if whether has one solution, no solution, or infinitely many solutions. show or explain your thinking.

a $3x - 7x + 1 = -4x + 5$ \t\tb $4(x + 3) = -2(2x + 6)$

1. elena said $4x - 6 = 4(x - 6)$ has infinitely many solutions. do you agree with elena’s answer? explain your thinking.

Explanation:

Problem 1

Step1: Simplify $x+2=x+4$

Subtract $x$ from both sides: $2=4$ (false)

Step2: Simplify $2x+4=2x+1+3$

Simplify right side: $2x+4=2x+4$ (always true)

Step3: Simplify $3(x-1)=2x+1$

Expand left side: $3x-3=2x+1$; subtract $2x$: $x-3=1$; add 3: $x=4$ (one solution)

Step4: Simplify $-(4x-5)=-x+5-3x$

Expand left side: $-4x+5=-4x+5$ (always true)

An equation that is never true for any $x$ cannot be satisfied by any value, so it has no solution.

Part a

Step1: Simplify left side

$3x-7x+1=-4x+1$

Step2: Set equal to right side

$-4x+1=-4x+5$; add $4x$: $1=5$ (false)

Part b

Step1: Expand both sides

Left: $4x+12$; Right: $-4x-12$

Step2: Solve for $x$

$4x+12=-4x-12$; add $4x$: $8x+12=-12$; subtract 12: $8x=-24$; divide by 8: $x=-3$

Answer:

Equation	One solution	No solution	Infinitely many solutions
$2x+4=2x+1+3$			✅
$3(x-1)=2x+1$	✅
$-(4x-5)=-x+5-3x$			✅

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Problem 2