C
Okay, so I need to solve this problem where I have to find the value of C in the equation 3C - 5 = 10. Hmm, let's see. Alright, first, I remember that to solve for a variable, I need to isolate it on one side of the equation. In this case, the variable is C. The equation is 3C minus 5 equals 10.
Let me write that down again to visualize it better: 3C - 5 = 10. So, the goal is to get C by itself. The first step should be to get rid of that -5 on the left side. To do that, I can add 5 to both sides of the equation. That way, I keep the equation balanced. Let me try that.
Adding 5 to both sides: 3C - 5 + 5 = 10 + 5. Simplifying both sides, the -5 and +5 on the left cancel each other out, so I have 3C = 15. Okay, that looks right. Now, the next step is to get rid of the coefficient 3 that's multiplied by C. To undo multiplication, I should divide both sides by 3. Let me do that.
Dividing both sides by 3: 3C / 3 = 15 / 3. Simplifying that, the 3's on the left cancel out, leaving C = 5. Wait, so is C equal to 5? Let me check my work to make sure I didn't make any mistakes.
Starting with the original equation: 3C - 5 = 10. If I substitute C with 5, does the equation hold true? Let's see. 3 times 5 is 15, and 15 minus 5 is 10. Yes, that equals 10, which matches the right side of the equation. So that checks out.
But just to be thorough, let me go through each step again. First, adding 5 to both sides: 3C -5 +5 becomes 3C, and 10 +5 becomes 15. So 3C =15. Then dividing both sides by 3: 3C/3 is C, and 15/3 is 5. So C=5. Yep, that seems correct.
https://www.18122.com/url/usdt/usdt1913.png
I guess another way to look at it is by breaking down the operations. The equation is saying that if you take a number C, multiply it by 3, and then subtract 5, you get 10. To reverse this process, you do the opposite operations in the reverse order. So, the last operation done was subtracting 5, so first add 5 to both sides. Then, the first operation was multiplying by 3, so divide by 3. That gives the same result.
Alternatively, someone might try to distribute or do something else, but in this case, the equation is straightforward. There's no need for distributing because there are no parentheses. It's a linear equation with one variable.
Wait, could there be a case where there are multiple solutions? For linear equations of the first degree, there's usually exactly one solution unless the equation is an identity or a contradiction. In this case, 3C -5 =10 is not an identity because it's only true for a specific C, and not for all C. So, only one solution exists, which is C=5.
Let me also consider if there's any possibility of division by zero or other undefined operations here. But since we're dealing with real numbers and the operations are basic arithmetic, there's no division by zero issue here.
Another check: if I graph the equation 3C -5 =10, it would be a vertical line at C=5 on the C-axis. But since this is a linear equation in one variable, the graph is just a point on the number line at 5.
Alternatively, creating a table of values might help. Let's see:
If C=4: 3*4 -5 =12-5=7, which is less than 10.
If C=5: 15-5=10, which matches.
If C=6: 18-5=13, which is more than 10.
So, yes, only when C is 5 does the equation equal 10.
I think that's solid. Maybe overcomplicating it, but better safe than sorry. So, the answer is C=5. Therefore, the value of C is 5.
To solve the equation (3C - 5 = 10) for (C):
Add 5 to both sides to isolate the term with (C):
[
3C - 5 + 5 = 10 + 5 \implies 3C = 15
]
Divide both sides by 3 to solve for (C):
[
\frac{3C}{3} = \frac{15}{3} \implies C = 5
]
Verification:
Substitute (C = 5) back into the original equation:
[
3(5) - 5 = 15 - 5 = 10
]
The left side equals the right side, confirming the solution is correct.
Answer: (\boxed{5})
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