Linear Equations in One Variable - Class 8 Mathematics - Chapter 2 - Notes, NCERT Solutions & Extra Questions
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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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Solve $\frac{6 x+1}{3}+1=\frac{x-3}{6}$
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Solve $5 x-2(2 x-7)=2(3 x-1)+\frac{7}{2}$
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Ask Chatterbot AIExercise 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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Ask Chatterbot AIExercise 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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Ask Chatterbot AIExtra 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.
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Let $y^{2}=\frac{(x+5)(x-3)}{x-1}$, then all the real values of $x$, for which $y$ has non-zero real values, are:
(A) $-5 < x < 1$
(B) $x > 15$
(C) $-5 < x < 1$ or $x > 3$
(D) None of these
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For the given linear equation y - 3x + 1 = 0, what are the values of y for x = {3, -4, 2}?
Option 1) 5, -8, 13
Option 2) 8, 5, 13
Option 3) 8, 8, 8
Option 4) 8, -13, 5
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