Question

1) 15*(2х*11)=285 2) 7*(123-4х)=749

Answer 1

1)
15(2х+11) = 258
30х+165 = 258
30х = 258-165
30х = 93
х = 93 / 30
х = 3.1
2)
7(123-4х)=749
123-4х=749\7
123-4х=107
4х=123-107
4х=16
х=16\4
х=4

Answer 2

It seems like you've provided two equations with variables and numerical values. It looks like you're looking for solutions to these equations. Let's solve them step by step:

Equation 1: $15 \cdot (2x \cdot 11) = 285$

First, simplify the expression inside the parentheses: $2x \cdot 11 = 22x$

Now substitute this back into the equation: $15 \cdot 22x = 285$

Divide both sides by 15: $22x = \frac{285}{15} = 19$

Finally, solve for $x$: $x = \frac{19}{22}$

Equation 2: $7 \cdot (123 - 4x) = 749$

Simplify the expression inside the parentheses: $123 - 4x = 123 - 4x$

Now substitute this back into the equation: $7 \cdot (123 - 4x) = 749$

Distribute the 7 on the left side: $861 - 28x = 749$

Subtract 861 from both sides: $-28x = 749 - 861 = -112$

Divide both sides by -28: $x = \frac{-112}{-28} = 4$

So, the solutions to the given equations are:

1. $x = \frac{19}{22}$
2. $x = 4$