Question
Solve the following equations. Check your result in case. $\frac{\text{y}+7}{3}=1+\frac{3\text{y}-2}{5}$
✓
Answer
$\frac{\text{y}+7}{3}=1+\frac{3\text{y}-2}{5}$
$\frac{5(\text{y}+7)=15+3(3\text{y}-2)}{15}$
($LCM$ of $3, 5 = 15)$ $5\text{y}+35=15+9\text{y}-6$
$5\text{y}-9\text{y}=15-6-25$ (By transposing) $-4\text{y}=-26$
$\text{y}=\frac{-26}{-4}=\frac{13}{2}$
$\therefore\text{y}=\frac{13}{2}$
Check: $\text{L.H.S.}=\frac{\text{y}-7}{3}=\frac{\frac{13}{2}+7}{3}$
$=\frac{13+14}{2\times3}$
$=\frac{27}{6}=\frac{9}{2}$
$\text{R.H.S.}=1+\frac{3\text{y}-2}{5}$
$=1+\frac{3\times\frac{13}{2}-2}{5}$
$=1+\frac{\frac{39}{2}-\frac{2}{1}}{5}$
$=1+\frac{39-4}{2\times5}$
$=1+\frac{35}{10}$
$=1+\frac{7}{2}=\frac{2+7}{2}=\frac{9}{2}$
$\therefore\text{L.H.S. = R.H.S.}$
Hence $\text{y}=\frac{13}{2}$
$\frac{5(\text{y}+7)=15+3(3\text{y}-2)}{15}$
($LCM$ of $3, 5 = 15)$ $5\text{y}+35=15+9\text{y}-6$
$5\text{y}-9\text{y}=15-6-25$ (By transposing) $-4\text{y}=-26$
$\text{y}=\frac{-26}{-4}=\frac{13}{2}$
$\therefore\text{y}=\frac{13}{2}$
Check: $\text{L.H.S.}=\frac{\text{y}-7}{3}=\frac{\frac{13}{2}+7}{3}$
$=\frac{13+14}{2\times3}$
$=\frac{27}{6}=\frac{9}{2}$
$\text{R.H.S.}=1+\frac{3\text{y}-2}{5}$
$=1+\frac{3\times\frac{13}{2}-2}{5}$
$=1+\frac{\frac{39}{2}-\frac{2}{1}}{5}$
$=1+\frac{39-4}{2\times5}$
$=1+\frac{35}{10}$
$=1+\frac{7}{2}=\frac{2+7}{2}=\frac{9}{2}$
$\therefore\text{L.H.S. = R.H.S.}$
Hence $\text{y}=\frac{13}{2}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Draw an $\angle\text{BAC}$ of measure $50^\circ $ such that $AB = 5cm$ and $AC = 7cm$. Through $C$ draw a line parallel to $AB$ and through $B$ draw a line parallel to $AC$, intersecting each other at $D$. Measure $BD$ and $CD.$
→Here are the shadows of some $3-D$ objects, when seen under the lamp of an Over$-$head projector. Identify the solid$(s)$ that match each shadow. $($There may be multiple answers for these$.)$
→Identify the given pair of lines as either parallel, perpendicular, or intersecting.
(a) Lines FQ and HT are ___________ lines.
(b) Lines FQ and MW are ___________ lines.
(c) Lines CN and FQ are ___________ lines.
(d) Lines DZ and YG are ___________ lines.
(e) Lines CN and YG are ___________ lines.
(f) Lines HT and MW are ___________lines.
(a) Lines FQ and HT are ___________ lines.
(b) Lines FQ and MW are ___________ lines.
(c) Lines CN and FQ are ___________ lines.
(d) Lines DZ and YG are ___________ lines.
(e) Lines CN and YG are ___________ lines.
(f) Lines HT and MW are ___________lines.
Separate positive and negative rational numbers from the following rational numbers:$\frac{-5}{-7}, \frac{12}{-5}, \frac{7}{4}, \frac{13}{-9}, 0, \frac{-18}{-7}, \frac{-95}{116}, \frac{-1}{-9}$
→Your friend found the median and the mode of a given data. Describe and correct your friends error if any $35,32,35,42,38,32,34$.
Median $=42$ and Mode $=32$
→Median $=42$ and Mode $=32$
The marks of a student in different subjects are given below:
Draw a bar graph from the above information.
→|
Subject
|
Hindi
|
English
|
Maths
|
Science
|
Social science
|
|
Marks
|
$43$
|
$56$
|
$80$
|
$65$
|
$50$
|
The product of two numbers is $25 \frac{5}{6}$. If one of the numbers is $6 \frac{2}{3}$, find the other,
→Rita has bought a carpet of size $4\text{m}\times6\frac{2}{3}\text{m}$ But her room size is $3\frac{1}{3}\text{m}\times5\frac{1}{3}\text{m}$ What fraction of area should be cut off to fit wall to wall carpet into the room?
→Draw a $\triangle\text{ABC}$ with $AB = 6cm, BC = 7cm$ and $CA = 8cm.$ Using ruler and compass alone, draw $(i)$ the bisector $AD$ of $\angle\text{A}$ and $(ii)$ perpendicular $AL$ from A on $BC$. Measure $LAD.$
→The production of saleable steel in some of the steel plants of our country during $1999$ is given below:
Construct a bar graph to represent the above data on a graph paper by using the scale $1$ big division$=20$ thousand tonnes.
→|
Plant
|
Bhilai
|
Durgapur
|
Rourkela
|
Bokaro
|
|
Production(in thousand tonnes)
|
$160$
|
$80$
|
$200$
|
$150$
|