There are many ways to start solving this equation. One way is to use the distributive property to rewrite the left side. This would eliminate the parentheses on the left so that we are closer to isolating c on the left. Start by distributing the 5 to the c. How can you represent 5·c?
5(c −4)
=
−15
−
=
−15
You got it!
Next, distribute the 5 to the 4. What is 5 • 4? We can subtract that from 5c.
5(c−4)
=
−15
5c −
=
−15
Great job!
Now that there aren't any parentheses on the left, let's think about getting the c term alone on that side. We can do that by moving all of the numbers to the right side. You can think of subtracting 20 as negative 20. What move could you make next to get rid of -20 on the left side, while keeping the equation balanced?
5(c−4)
=
−15
5c − 20
=
−15
Multiply both sides by 20
Add 20 to both sides
Subtract 20 from both sides
Good work!
Show that you can add 20 on both sides of the equation to get all of the numbers to the right, while keeping the equation balanced.
5(c−4)
=
−15
5c−20
=
−15
+
20
+
20
You got it!
Add 20 on both sides and rewrite the equation to show the values that remain.
5(c−4)
=
−15
5c−20
=
−15
+
20
+
20
=
Great job!
Now we have 5c alone on the left, but c isn't alone yet. What balanced move can we make next to get c alone on the left?
5(c−4)
=
−15
5c−20
=
−15
+
20
+
20
5c
=
5
Multiply by 5
Divide by 5
Subtract 5c
You got it!
Show that you can divide both sides by 5 to get c alone on the left and keep the equation balanced.
5(c−4)
=
−15
5c−20
=
−15
+
20
+
20
5c
=
5
5
5
Excellent!
Divide both sides by 5 and rewrite the equation to solve for the value of c.
5(c−4)
=
−15
5c−20
=
−15
+
20
+
20
5c
=
5
5
5
1 answer
Let's go through the steps you've outlined to solve for \( c \) in the equation \( 5(c - 4) = -15 \).
Distributing the 5: \[ 5(c - 4) = 5c - 20 \] So the equation becomes: \[ 5c - 20 = -15 \]
Adding 20 to both sides: \[ 5c - 20 + 20 = -15 + 20 \] Which simplifies to: \[ 5c = 5 \]
Dividing both sides by 5: \[ \frac{5c}{5} = \frac{5}{5} \] This yields: \[ c = 1 \]
Thus, the value of \( c \) is \( \boxed{1} \).
