Answer :

To solve the equation [tex]\(\frac{2x}{3} + 1 = \frac{7x}{15} + 3\)[/tex], we will follow a step-by-step approach:

### Step 1: Clear the fractions

To eliminate the fractions, we need to find a common denominator. The common denominator for both 3 and 15 is 15. We will now multiply every term of the equation by 15 to clear these fractions:

[tex]\[ 15 \left(\frac{2x}{3}\right) + 15 \cdot 1 = 15 \left(\frac{7x}{15}\right) + 15 \cdot 3 \][/tex]

### Step 2: Simplify the equation

Now we simplify each term:

[tex]\[ 5 \cdot 2x + 15 = 7x + 45 \][/tex]

This simplifies further to:

[tex]\[ 10x + 15 = 7x + 45 \][/tex]

### Step 3: Get all the [tex]\(x\)[/tex] terms on one side and constants on the other

To isolate [tex]\(x\)[/tex], we subtract [tex]\(7x\)[/tex] from both sides of the equation:

[tex]\[ 10x - 7x + 15 = 45 \][/tex]

This simplifies to:

[tex]\[ 3x + 15 = 45 \][/tex]

We then subtract 15 from both sides:

[tex]\[ 3x = 30 \][/tex]

### Step 4: Solve for [tex]\(x\)[/tex]

Finally, divide both sides by 3 to solve for [tex]\(x\)[/tex]:

[tex]\[ x = \frac{30}{3} \][/tex]

So:

[tex]\[ x = 10 \][/tex]

### Verifying the solution

Let's verify the solution by plugging [tex]\(x = 10\)[/tex] back into the original equation:

[tex]\[ \frac{2(10)}{3} + 1 = \frac{7(10)}{15} + 3 \][/tex]

Calculate each side separately:

[tex]\[ \frac{20}{3} + 1 = \frac{70}{15} + 3 \][/tex]

Simplify [tex]\(\frac{70}{15}\)[/tex] to [tex]\(\frac{14}{3}\)[/tex]:

[tex]\[ \frac{20}{3} + 1 = \frac{14}{3} + 3 \][/tex]

Convert 3 to a fraction with a denominator of 3:

[tex]\[ \frac{20}{3} + 1 = \frac{14}{3} + \frac{9}{3} \][/tex]

This simplifies to:

[tex]\[ \frac{20}{3} + 1 = \frac{23}{3} \][/tex]

Now add 1 on the left side (express 1 as [tex]\(\frac{3}{3}\)[/tex]):

[tex]\[ \frac{20}{3} + \frac{3}{3} = \frac{23}{3} \][/tex]

Both sides of the equation are equal, thus confirming our solution is correct.

In conclusion, the solution to the equation [tex]\(\frac{2x}{3} + 1 = \frac{7x}{15} + 3\)[/tex] is:

[tex]\[ x = 10 \][/tex]