Ncert solutions

Ncert solutions

Grade 8

Linear Equations in One Variable | Exercise 2.5

Question 1

Q1) Solve the following linear equations.

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

  • Solution

  • Transcript

Solution:

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

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

\Rightarrow\ \frac{3x-2x}{6}=\frac{5+4}{20}

\Rightarrow\ \frac{x}{6}=\frac{9}{20}

\Rightarrow x=\frac{9\times6}{20}=\frac{27}{10}

To check:

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

\Rightarrow\ \frac{27}{10\times2}-\frac{1}{5}=\frac{27}{10\times3}+\frac{1}{4}

\Rightarrow\ \frac{27}{20}-\frac{1}{5}=\frac{9}{10}+\frac{1}{4}

\Rightarrow\ \frac{27-4}{20}=\frac{18+5}{20}

\Rightarrow\ \frac{23}{20}=\frac{23}{20}

L.H.S. = R. H. S.

Therefore, it is correct.

hi children our today's question is solve the following linear equations x by 2 minus 1 by 5 is equal to x by 3 plus 1 by 4. now whenever I have such questions the first thing that I need to make sure is that all of my preferred terms are on the same side of the equality sign and hence I will be transposing this x by 3 to the left-hand side and this minus 1 by 5 to the right-hand side In doing so I get x by 2 minus x by 3 is equal to 1 by 4 plus 1 by 5. now also kids observe the way that i have changed the signs of the terms that I have transposed now on finding just cross multiplying the terms here so cross multiplying this I get 3 into x which is 3x minus 2 into x that is 2x upon 3 into 2 that is 6 which is equal to again now 5 into 1 so 5 plus 4 into 1 that is 4 upon 4 multiplied by 5 which is 20. so now I get 3x minus 2x which is x by 6 is equal to 5 plus 4 that is 9 upon 20 so now I can transpose this x from here to the numerator of the right-hand side which then becomes x is equal to 9 into 6 by 20 which is x is equal to 9 sigmas 54 by 20 and then the final answer that i get is x equal to 2.7

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Question 1

Q1) Solve the following linear equations.

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

  • Solution

  • Transcript

Solution:

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

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

\Rightarrow\ \frac{3x-2x}{6}=\frac{5+4}{20}

\Rightarrow\ \frac{x}{6}=\frac{9}{20}

\Rightarrow x=\frac{9\times6}{20}=\frac{27}{10}

To check:

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

\Rightarrow\ \frac{27}{10\times2}-\frac{1}{5}=\frac{27}{10\times3}+\frac{1}{4}

\Rightarrow\ \frac{27}{20}-\frac{1}{5}=\frac{9}{10}+\frac{1}{4}

\Rightarrow\ \frac{27-4}{20}=\frac{18+5}{20}

\Rightarrow\ \frac{23}{20}=\frac{23}{20}

L.H.S. = R. H. S.

Therefore, it is correct.

hi children our today's question is solve the following linear equations x by 2 minus 1 by 5 is equal to x by 3 plus 1 by 4. now whenever I have such questions the first thing that I need to make sure is that all of my preferred terms are on the same side of the equality sign and hence I will be transposing this x by 3 to the left-hand side and this minus 1 by 5 to the right-hand side In doing so I get x by 2 minus x by 3 is equal to 1 by 4 plus 1 by 5. now also kids observe the way that i have changed the signs of the terms that I have transposed now on finding just cross multiplying the terms here so cross multiplying this I get 3 into x which is 3x minus 2 into x that is 2x upon 3 into 2 that is 6 which is equal to again now 5 into 1 so 5 plus 4 into 1 that is 4 upon 4 multiplied by 5 which is 20. so now I get 3x minus 2x which is x by 6 is equal to 5 plus 4 that is 9 upon 20 so now I can transpose this x from here to the numerator of the right-hand side which then becomes x is equal to 9 into 6 by 20 which is x is equal to 9 sigmas 54 by 20 and then the final answer that i get is x equal to 2.7

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

Learn from an expert tutor.

Book a free class now
subject-cta
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