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Solve for [tex]\( x \)[/tex]:

1. [tex]\( 3x + 10 = 40 \)[/tex]

2. [tex]\( 4A - C = 10A + 5 \)[/tex]

Solve for [tex]\( y \)[/tex]:

3. [tex]\(\frac{M - 5B}{C^3} - 8 = 10\)[/tex]

4. [tex]\(\frac{\sqrt{4y} - 10}{2T} = F - 2T \frac{\sqrt{4y - 10}}{F}\)[/tex]


Sagot :

To find the required values, we can solve the given equations step-by-step.

Step-by-Step Solution

### Solving for [tex]\( x \)[/tex] in Equation (1):

Equation (1): [tex]\( 3x + 10 = 40 \)[/tex]

1. Subtract 10 from both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ 3x = 40 - 10 \][/tex]

2. Simplify the right-hand side:
[tex]\[ 3x = 30 \][/tex]

3. Divide both sides by 3 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{30}{3} = 10 \][/tex]

So, [tex]\( x = 10 \)[/tex].

### Solving for [tex]\( C \)[/tex] in Equation (2):

Equation (2): [tex]\( 4A - C = 10A + 5 \)[/tex]

1. To solve for [tex]\( C \)[/tex], subtract [tex]\( 4A \)[/tex] from both sides:
[tex]\[ -C = 10A + 5 - 4A \][/tex]

2. Simplify the right-hand side:
[tex]\[ -C = 6A + 5 \][/tex]

3. Multiply both sides by -1 to isolate [tex]\( C \)[/tex]:
[tex]\[ C = -6A - 5 \][/tex]

We simplify it to [tex]\( C = 6A + 5 - 4A \)[/tex], hence:
[tex]\[ C = 6A + 5 \][/tex]

So, [tex]\( C = 6A + 5 \)[/tex].

### Solving for [tex]\( y \)[/tex] in Equation (4):

Equation (4): [tex]\(\frac{M - 5B}{C^3} - 8 = 10\)[/tex]

1. Add 8 to both sides to move the -8 to the right-hand side:
[tex]\[ \frac{M - 5B}{C^3} = 18 \][/tex]

2. Multiply both sides by [tex]\( C^3 \)[/tex] to clear the fraction:
[tex]\[ M - 5B = 18C^3 \][/tex]

3. Solve for [tex]\( y \)[/tex]:
[tex]\[ y = 18C^3 + 5B + M \][/tex]

So, [tex]\( y = 18C^3 + 5B + M \)[/tex].

Summary of the solutions:
- [tex]\( x = 10 \)[/tex]
- [tex]\( C = 6A + 5 \)[/tex]
- [tex]\( y = 18C^3 + 5B + M \)[/tex]

The problem description mentions Equation (5), but without additional context or full information, we're unable to determine the exact methodology needed to solve for [tex]\( y \)[/tex] in that case. Therefore, the interpretation follows from the previously established results.