# Linear Equations in One Variable - Class 8 - Mathematics

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## Examples - Linear Equations in One Variable | NCERT | Mathematics | Class 8

Solve $2 x-3=x+2$

The solution to the equation $2x - 3 = x + 2$ is $x = 5$.

Here is the plot showing the lines $2x - 3$ and $x + 2$, illustrating the point of intersection:

And here is the depiction on the number line where the solution $x = 5$ is located:

Solve $5 x+\frac{7}{2}=\frac{3}{2} x-14$

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Sign up nowSolve $\frac{6 x+1}{3}+1=\frac{x-3}{6}$

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Sign up nowSolve $5 x-2(2 x-7)=2(3 x-1)+\frac{7}{2}$

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Sign up now## Exercise 2.1 - Linear Equations in One Variable | NCERT | Mathematics | Class 8

$3 x=2 x+18$

The solution to the equation (3x = 2x + 18) is (x = 18).

You can visualize the solution on a plot and number line below:

Plot:

Number Line:

This means that (x) must be 18 for the equation to hold true.

$5 t-3=3 t-5$

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$5 x+9=5+3 x$

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$4 z+3=6+2 z$

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$2 x-1=14-x$

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$8 x+4=3(x-1)+7$

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$x=\frac{4}{5}(x+10)$

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$\frac{2 x}{3}+1=\frac{7 x}{15}+3$

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$2 y+\frac{5}{3}=\frac{26}{3}-y$

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$3 m=5 m-\frac{8}{5}$

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## Exercise 2.2 - Linear Equations in One Variable | NCERT | Mathematics | Class 8

$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

To solve the equation $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$, we'll follow these steps:

Find a common denominator for the fractions to simplify the equation.

Combine like terms to isolate variables on one side and constants on the other.

Solve for (x).

Let's solve it:

The equation is: $$ \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4} $$

First, we'll multiply both sides of the equation by the common denominator of 2, 3, 4, and 5, which is 60, to eliminate the fractions: $$ 60 \cdot \left( \frac{x}{2}-\frac{1}{5} \right) = 60 \cdot \left( \frac{x}{3}+\frac{1}{4} \right) $$

Simplifying, we get: $$ 30x - 12 = 20x + 15 $$

Subtracting (20x) from both sides and adding (12) to both sides to isolate (x), we obtain: $$ 30x - 20x = 15 + 12 $$

Therefore, we simplify to find (x): $$ 10x = 27 $$

Finally, divide by (10) to solve for (x): $$ x = \frac{27}{10} $$

Thus, the solution to the equation is (x = \frac{27}{10}).

$\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21$

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$x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}$

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$\frac{x-5}{3}=\frac{x-3}{5}$

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$\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t$

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$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

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$3(t-3)=5(2 t+1)$

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$15(y-4)-2(y-9)+5(y+6)=0$

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$3(5 z-7)-2(9 z-11)=4(8 z-13)-17$

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$0.25(4 f-3)=0.05(10 f-9)$

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## Extra Questions - Linear Equations in One Variable | NCERT | Mathematics | Class 8

If $\frac{3x+5}{4} + 1 = \frac{x+3}{6}$, then find the value of $x$.

A) 3

B) -3

C) 6

D) -6

The correct option is **B) -3**

**Given Equation:**

$$ \frac{3x+5}{4} + 1 = \frac{x+3}{6} $$

**Step 1: Add 1 to the numerator on the left-hand side:**

$$ \frac{3x + 5 + 4}{4} = \frac{x+3}{6} $$ $$ \frac{3x + 9}{4} = \frac{x+3}{6} $$

**Step 2: Cross-multiply to eliminate the fractions:**

$$ 6(3x + 9) = 4(x + 3) $$ $$ 18x + 54 = 4x + 12 $$

**Step 3: Solve for (x):**

$$ 18x - 4x = 12 - 54 $$ $$ 14x = -42 $$ $$ x = \frac{-42}{14} $$ $$ x = -3 $$

Thus, (x) equals **-3**.

You deposit ₹345 into a bank account at 7% simple interest per year. How much interest would you earn after 2 years

A) ₹48.30

B) ₹48.15

C) ₹48.10

D) ₹47.20"

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Select the variable in the following expression: $$ y - 14 + 144 $$

A) $y - 14$

B) $144$

C) $14$

D) $y$