Question

x-1/3=x+2/6+3
linear equations​

Answer 1

Answer:

SOLUTION

Answer:

see this

Step-by-step explanation:

How to solve your question

Your question is

−

1

⋅

1

3

=

+

2

6

+

3

x-1 \cdot \frac{1}{3}=x+\frac{2}{6}+3

x−1⋅31=x+62+3

Solve

1

Multiply the numbers

−

1

⋅

1

3

=

+

2

6

+

3

x{\color{#c92786}{-1}} \cdot {\color{#c92786}{\frac{1}{3}}}=x+\frac{2}{6}+3

x−1⋅31=x+62+3

−

1

3

=

+

2

6

+

3

x{\color{#c92786}{-\frac{1}{3}}}=x+\frac{2}{6}+3

x−31=x+62+3

2

Divide the numbers

3

Add the numbers

4

Add

1

3

\frac{1}{3}

31

to both sides of the equation

5

Simplify

Step-by-step explanation:

hope it helps

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Answer 2

Answer:

∴ The required value of $x$ is $2$.

Step-by-step explanation:

In context to question asked,

• We have to find the value of $x$.
• The given equation is,

                $\frac{x-1}{3} =\frac{x+2}{6+3}$

             ∴ $\frac{x-1}{3} = \frac{x+2}{9}$

• By cross - multiplication, we get

       ∴ $9(x-1)=3(x+2)$

• Opening the brackets, we get

          ∴ $9x-9=3x+6$

• Collect the terms with variable at one side and constant terms at another side, we get

          ∴ $9x-3x=6+9$

                 ∴ $6x=12$

• Transpose $6$ at other side, we get

                   ∴ $x=\frac{12}{6}$

                   ∴ $x=2$

• ∴ The required value of $x$ is $2$.