Determine if each equation has exactly 1 solution, zero solutions or infinite solutions.

Question

ROW1
Exactly 1 Solution

Zero Solutions

Infinite Solutions

Drag & Drop the answer
7x+5=7x−57x+5=7x-57x+5=7x−5

3x+6=−3x−63x+6=-3x-63x+6=−3x−6

3x+6=6+3x3x+6=6+3x3x+6=6+3x

3x+6=3x+73x+6=3x+73x+6=3x+7

Answer 1

To determine the number of solutions for each equation, let's analyze them one by one.

1. Equation: \(7x + 5 = 7x - 5\)

   Subtract \(7x\) from both sides: \[ 5 = -5 \] This statement is false, indicating that there are no solutions.

   • Zero Solutions

2. Equation: \(3x + 6 = -3x - 6\)

   Adding \(3x\) to each side gives: \[ 6 + 6 = -3x + 3x \implies 12 = 0 \] This is also false, so there are no solutions.

   • Zero Solutions

3. Equation: \(3x + 6 = 6 + 3x\)

   Subtracting \(3x\) from both sides results in: \[ 6 = 6 \] This is a true statement and holds for all values of \(x\), indicating that there are infinitely many solutions.

   • Infinite Solutions

4. Equation: \(3x + 6 = 3x + 7\)

   Subtracting \(3x\) from both sides results in: \[ 6 = 7 \] This statement is false, indicating no solutions.

   • Zero Solutions

Final Summary:

• \(7x + 5 = 7x - 5\): Zero Solutions
• \(3x + 6 = -3x - 6\): Zero Solutions
• \(3x + 6 = 6 + 3x\): Infinite Solutions
• \(3x + 6 = 3x + 7\): Zero Solutions

Answer 2

Solve the following two-step linear equation.

−
2
x
+
20
4
=
7
4
−2x+20
​
=7



x
=
x=

Answer 3

To solve the equation \(-2x + 20 = 7\), we will follow these steps:

1. Isolate the term with \(x\):

   \[-2x + 20 = 7\]

   Subtract 20 from both sides:

   \[-2x = 7 - 20\]

   \[-2x = -13\]

2. Solve for \(x\):

   Divide both sides by -2:

   \[x = \frac{-13}{-2}\]

   \[x = \frac{13}{2} \]

Thus, the solution is:

\[ x = \frac{13}{2} \] or 6.5 in decimal form.

Answer 4

Solve the following linear equation.

2z+10+6z+14=−562z+10+6z+14=-562z+10+6z+14=−56

z=z=z=

Answer 5

To solve the equation \(2z + 10 + 6z + 14 = -56\), we will follow these steps:

1. Combine like terms on the left side:

   \[ 2z + 6z + 10 + 14 = -56 \]

   Combine \(2z\) and \(6z\):

   \[ (2z + 6z) + (10 + 14) = -56 \]

   Which simplifies to:

   \[ 8z + 24 = -56 \]

2. Isolate the term with \(z\):

   Subtract 24 from both sides:

   \[ 8z = -56 - 24 \]

   \[ 8z = -80 \]

3. Solve for \(z\):

   Divide both sides by 8:

   \[ z = \frac{-80}{8} \]

   \[ z = -10 \]

Thus, the solution is:

\[ z = -10 \]

Answer 6

Finish the equation so that the equation has infinite amount of solutions. Click Here for Rubric

4
(
6
x
+
2
)
−
6
=
4(6x+2)−6=




Finish the equation so that the equation that has zero solutions

5
(
6
x
+
7
)
−
6
=
5(6x+7)−6=



x
+
x+


Find the value of
x
x that makes the following have exactly 1 solution:

5
(
4
x
+
8
)
=
16
x
+
4
5(4x+8)=16x+4

x
=
x=

Answer 7

Let's break down the three tasks you provided one by one.

1. Equation with Infinite Solutions

The original equation is: \[ 4(6x + 2) - 6 = \] To achieve infinite solutions, you can set the left side equal to itself, which means you can multiply through by a non-zero constant. Here’s an example:

\[ 4(6x + 2) - 6 = 4(6x + 2) - 6 \]

This is always true, so it has infinite solutions.

2. Equation with Zero Solutions

The given equation is: \[ 5(6x + 7) - 6 = \] To create an equation with zero solutions, you can set the left side equal to a false statement. Here's an example:

\[ 5(6x + 7) - 6 = 10 \]

Now, if you compute the left side: \[ 5(6x + 7) - 6 = 30x + 35 - 6 = 30x + 29 \]

Now set it to equal 10: \[ 30x + 29 = 10 \]

This results in \(30x = 10 - 29 = -19\), which means you can solve for \(x\). However, if we adjust it slightly to: \[ 5(6x + 7) - 6 = 5(6x + 7) - 2 \] This leads to no solutions since the two sides are not equal.

3. Finding \( x \) for Exactly 1 Solution

The equation given is: \[ 5(4x + 8) = 16x + 4 \] First, expand the left side: \[ 20x + 40 = 16x + 4 \]

Now, isolate \(x\): \[ 20x - 16x = 4 - 40 \] \[ 4x = -36 \] \[ x = -9 \]

So the value of \(x\) that makes the equation have exactly one solution is: \[ x = -9 \]

Summary of Answers

1. Infinite solutions: \[ 4(6x + 2) - 6 = 4(6x + 2) - 6 \]
2. Zero solutions: \[ 5(6x + 7) - 6 = 10 \]
3. Value of \(x\) with exactly 1 solution: \[ x = -9 \]