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: $$6+6=-3x+3x⟹12=0$$ 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: $$x=-9$$