Question

A)4(2x-1)-3(2x-7)=17+2x б)6(3-5у)+12(у-3)18+18у=0в)9(2z-1)+5(3z+1)=3(11z-1)-1г)7(4t+5)-8(4-5t)-3-68t=0

Answer 1

Думаю, что во втором вы потеряли знак, на фотографии я поставил знак вопроса, где допущена ошибка

Answer 2

Sure, I'd be happy to help you solve these equations. Let's go through each one step by step:

Equation A: $4(2x-1)-3(2x-7)=17+2x$

1. Distribute the constants: $8x - 4 - 6x + 21 = 17 + 2x$
2. Combine like terms on both sides: $2x + 17 = 17 + 2x$
3. Subtract $2x$ from both sides: $17 = 17$

The equation simplifies to $17 = 17$, which is a true statement. This means that the equation is an identity, and it holds true for all values of $x$.

Equation B: $6(3-5y)+12(y-3)18+18y=0$

1. Distribute the constants: $18 - 30y + 12y - 36 + 18 + 18y = 0$
2. Combine like terms on both sides: $-30y + 12y + 18y = -18 - 18$
3. Combine the constants: $0 = -36$

The equation simplifies to $0 = -36$, which is not true. This means that the equation has no solution, as there is no value of $y$ that can satisfy it.

Equation C: $9(2z-1)+5(3z+1)=3(11z-1)-1$

1. Distribute the constants on the left side: $18z - 9 + 15z + 5 = 33z - 3 - 1$
2. Combine like terms on both sides: $33z - 4 = 33z - 4$

The equation simplifies to $33z - 4 = 33z - 4$, which is a true statement. This means that the equation is an identity and holds true for all values of $z$.

Equation D: $7(4t+5)-8(4-5t)-3-68t=0$

1. Distribute the constants: $28t + 35 - 32 + 40t - 3 - 68t = 0$
2. Combine like terms on both sides: $28t + 40t - 68t = 0 - 35 + 32 + 3$
3. Combine the constants: $0 = 0$

The equation simplifies to $0 = 0$, which is a true statement. This means that the equation is an identity and holds true for all values of $t$.

To summarize:

• Equation A is an identity and holds true for all values of $x$.
• Equation B has no solution; there is no value of $y$ that satisfies it.
• Equation C is an identity and holds true for all values of $z$.
• Equation D is an identity and holds true for all values of $t$.