Question

EXERCI
Solve each of the following equations and check the
result
⅔x + 1=7/3​

Answer 1

Step-by-step explanation:

The given equation is:

$\quad\frac{2}{3}x+1=\frac{7}{3}$

Subtract $1$ from both sides:

$\quad\frac{2}{3}x+1-1=\frac{7}{3}-1$

$\Rightarrow \frac{2}{3}x=\frac{7-3}{3}$

$\Rightarrow \frac{2}{3}x=\frac{4}{3}$

Multiply $\frac{3}{2}$ to both sides:

$\quad \frac{2}{3}\times\frac{3}{2}x=\frac{4}{3}\times\frac{3}{2}$

$\Rightarrow x=2$

Thus the required solution is $x=2$.

Check step:

Put $x=2$ to the left hand side of the given equation:

$\quad\frac{2}{3}\times 2+1$

$=\frac{4}{3}+1$

$=\frac{4+3}{3}$

$=\frac{7}{3}$

This satisfies the given equation.

Answer: required solution is $\mathrm{x=2}$.

Answer 2

Answer:

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The given equestion is:

⟹$\frac{2}{3} x + 1 = \frac{7}{3}$

Subtract 1 from both sides

⟹$\frac{2}{3} x \cancel+ 1 \cancel - 1 = \frac{7}{3} - 1 \\ \frac{2}{3} x = \frac{7 - 3}{3} \\⟹\frac{2}{3} x = \frac{4}{3}$

Multiply 3/2 in both sides

$\frac{ \cancel2}{ \cancel3} \times \frac{ \cancel3}{ \cancel2} x = \frac{ \cancel4 ^{2} }{ \cancel3} \times \frac{ \cancel3}{ \cancel2} \\ \orange\star\blue{\boxed {x = 2}} \orange \star$

Check

Put x=2 in LHS

$\frac{2}{3} x + 1 \\ \rightarrow \frac{2}{3} \times 2 + 1 \\\rightarrow \: \frac{4}{3} + 1 \\ \rightarrow \frac{4 + 3}{3} \\ \rightarrow \frac{7}{3} = RHS$

The required solution = 2

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