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Okay, so I need to figure out how to solve this problem where I have to find the value of x in the equation 3x + 5 = 20. Alright, let's start by recalling what I know about algebra. The goal here is to isolate x, right? That means I need to get x by itself on one side of the equation.
First, the equation is 3x plus 5 equals 20. So, 3x + 5 = 20. To get rid of that 5 on the left side, I should subtract 5 from both sides. That way, I keep the equation balanced. Let me write that down: 3x + 5 - 5 = 20 - 5. Simplifying both sides, the +5 and -5 on the left cancel each other out, so I get 3x = 15.
Now, I need to solve for x. Since 3 is multiplied by x, I should do the opposite operation, which is division. So, I divide both sides by 3. That would look like 3x/3 = 15/3. The 3s on the left cancel out, leaving x = 5.
Wait, let me check if that's correct. If I plug x = 5 back into the original equation, does it hold true? Let's see: 3*(5) + 5 = 15 + 5 = 20. Yes, that equals 20, which matches the right side of the equation. So, x = 5 is the correct solution.
But just to make sure I didn't make any mistakes in my steps, let me go through it again. Starting with 3x + 5 = 20. Subtract 5 from both sides: 3x = 15. Then divide by 3: x = 5. Yep, that seems right.
Another way to think about it is to reverse the operations. The last operation done on x is adding 5, but actually, the equation is 3 times x plus 5. So, to reverse it, first subtract 5, then divide by 3. That's exactly what I did.
I guess there's not much more to it. It's a straightforward linear equation. Maybe if there were variables on both sides or fractions, it would be more complex, but here it's pretty simple. Just two steps: subtract 5 and then divide by 3.
I wonder if there's an alternative method. For example, could I divide everything by 3 first? Let's try that. If I divide each term by 3, the equation becomes x + 5/3 = 20/3. Then, subtract 5/3 from both sides: x = 20/3 - 5/3 = 15/3 = 5. Oh, that works too. So, either method is valid. Whether you subtract first or divide first, you still end up with x = 5.
But usually, the standard approach is to reverse the order of operations, which is why subtracting 5 first is more intuitive. Dividing first might be less common but still mathematically correct. Either way, the answer is the same.
Just to ensure I haven't overlooked anything, let me consider if there are any special cases or if the equation could have multiple solutions. But since it's a linear equation with degree 1, there should be exactly one solution. So, x = 5 is the only answer.
No, I think that's solid. The check confirms it, and the alternative method leads to the same result. I don't see any errors in my reasoning. It's a simple problem, but it's good to verify each step to build good habits, especially for more complicated equations later on.
To solve the equation (3x + 5 = 20) for (x), follow these steps:
Subtract 5 from both sides to isolate the term with (x):
[
3x + 5 - 5 = 20 - 5 \implies 3x = 15
]
Divide both sides by 3 to solve for (x):
[
\frac{3x}{3} = \frac{15}{3} \implies x = 5
]
Verification:
Substitute (x = 5) back into the original equation:
[
3(5) + 5 = 15 + 5 = 20 \quad \text{(Which matches the right side of the equation)}
]
Answer: (\boxed{5})
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发表于 2025-11-6 09:34:42