1. ) [tex]$x+142=16 x+7$[/tex]
2. ) [tex]$8 x=2 x-6$[/tex]
3. ) [tex]$1+9 x=7+8 x$[/tex]
4. ) [tex]$-8 x+8=10+5-9 x$[/tex]
5. ) [tex]$7 x+2=-10 x+2$[/tex]
6. ) [tex]$55-x=2 x+13$[/tex]

Solve each linear equation by isolating x through algebraic manipulations.
Equation 1: x + 142 = 16 x + 7 leads to x = 9 .
Equation 2: 8 x = 2 x − 6 leads to x = − 1 .
Equation 3: 1 + 9 x = 7 + 8 x leads to x = 6 .
Equation 4: − 8 x + 8 = 10 + 5 − 9 x leads to x = 7 .
Equation 5: 7 x + 2 = − 10 x + 2 leads to x = 0 .
Equation 6: 55 − x = 2 x + 13 leads to x = 14 .
The solutions are: x = 9 , − 1 , 6 , 7 , 0 , 14 ​ .

Explanation

Problem Analysis We are given six linear equations, and our goal is to solve each one for the variable x . We will use algebraic manipulations to isolate x on one side of each equation.

Solving Equation 1



x + 142 = 16 x + 7 Subtract x from both sides: 142 = 15 x + 7 Subtract 7 from both sides: 135 = 15 x Divide both sides by 15: x = 15 135 ​ = 9 Therefore, x = 9 .


Solving Equation 2


8 x = 2 x − 6 Subtract 2 x from both sides: 6 x = − 6 Divide both sides by 6: x = 6 − 6 ​ = − 1 Therefore, x = − 1 .


Solving Equation 3


1 + 9 x = 7 + 8 x Subtract 8 x from both sides: 1 + x = 7 Subtract 1 from both sides: x = 6 Therefore, x = 6 .


Solving Equation 4


− 8 x + 8 = 10 + 5 − 9 x − 8 x + 8 = 15 − 9 x Add 9 x to both sides: x + 8 = 15 Subtract 8 from both sides: x = 7 Therefore, x = 7 .


Solving Equation 5


7 x + 2 = − 10 x + 2 Add 10 x to both sides: 17 x + 2 = 2 Subtract 2 from both sides: 17 x = 0 Divide both sides by 17: x = 17 0 ​ = 0 Therefore, x = 0 .


Solving Equation 6


55 − x = 2 x + 13 Add x to both sides: 55 = 3 x + 13 Subtract 13 from both sides: 42 = 3 x Divide both sides by 3: x = 3 42 ​ = 14 Therefore, x = 14 .


Final Answers In summary, the solutions to the six equations are:


x = 9
x = − 1
x = 6
x = 7
x = 0
x = 14

Examples
Imagine you're managing a budget for a school event. You have a certain amount of money to spend, and you need to figure out how many tickets you can sell to break even. Each equation represents a different scenario with varying costs and ticket prices. Solving these linear equations helps you determine the exact number of tickets needed to cover all expenses and make the event a success. This skill is crucial for making informed financial decisions in everyday life.

Answered by GinnyAnswer | 2025-07-08

The solutions for the six linear equations are x = 9, x = -1, x = 6, x = 7, x = 0, and x = 14. Each equation was solved by isolating the variable x through algebraic manipulations. Thus, the solutions have been found for each specific equation presented.
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Answered by Anonymous | 2025-07-10