Question
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$\frac{1}{5}(x-2)=\frac{1}{4}(1-y); 26 x+3 y+4=0$
✓
Answer
$ \frac{1}{5}(x-2)=\frac{1}{4}(1-y)$
$ \Rightarrow 4(x-2)=5(1-y)$
$ \Rightarrow 4 x-8=5-5 y$
$\Rightarrow 4 x+5 y=13\dots.....(1)$
$26 x+3 y=-4\dots.....(2)$
Multiplying equation no. $(1)$ by $3$ and $(2)$ by $5 .$
$12 x+15 y=39\dots.....(3)$
$130 x+15 y=-20\dots.....(4)$
Subtracting equation $(4)$ from $(3)$
$12 x+15 y =39$
$-130 x+15 y =-20$
$- +$
$-118 x =59$
$x =-\frac{59}{118}$
$x =-\frac{1}{2}$
From $(1)$
$4\left(-\frac{1}{2}\right)+5 y=13$
$ 5 y=13+2$
$ y=3$
$ \Rightarrow 4(x-2)=5(1-y)$
$ \Rightarrow 4 x-8=5-5 y$
$\Rightarrow 4 x+5 y=13\dots.....(1)$
$26 x+3 y=-4\dots.....(2)$
Multiplying equation no. $(1)$ by $3$ and $(2)$ by $5 .$
$12 x+15 y=39\dots.....(3)$
$130 x+15 y=-20\dots.....(4)$
Subtracting equation $(4)$ from $(3)$
$12 x+15 y =39$
$-130 x+15 y =-20$
$- +$
$-118 x =59$
$x =-\frac{59}{118}$
$x =-\frac{1}{2}$
From $(1)$
$4\left(-\frac{1}{2}\right)+5 y=13$
$ 5 y=13+2$
$ y=3$
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
A kite is flying at a height of 120 m from the level ground. It is attached to a string inclined at 60 to the horizontal. Find the length of the string. (Take $\sqrt{3}=1.73$.)
→Given below are the marks obtained by $30$ students in an examination:
Taking class intervals $1 - 10, 11 - 20,,\dots ....., 41 - 50;$make a frequency table for the above distribution.
→| $08$ | $17$ | $33$ | $41$ | $47$ | $23$ | $20$ | $34$ |
| $09$ | $18$ | $42$ | $14$ | $30$ | $19$ | $29$ | $11$ |
| $36$ | $48$ | $40$ | $24$ | $22$ | $02$ | $16$ | $21$ |
| $15$ | $32$ | $47$ | $44$ | $33$ | $01$ |
Draw parallelogram $\text{ABCD}$ with the following data:$A B=6 \ cm , A D=5 \ cm$ and $\angle D A B=45^{\circ}$.Let $A C$ and $D B$ meet in $O$ and let $E$ be the mid$-$point of $B C$. Join $O E$.Prove that:(i) $OE \| AB;(ii)OE =\frac{1}{2} AB$.
→In the given figure, $A B=B C$ and $A C=C D$. Show that : $\angle BAD : \angle ADB =3: 1$
→Case Study : In 2015, the population of a town was 20000. During 2016 and 2017, it increases by 4% and 5% every year respectively. In 2018, due to an epidemic and migration to cities, the population decreases at 5%. Based on the above information answer the following questions:
1. The population of the town in 2016 was:
(a) 20800 (b) 20500 (c) 20460 (d) 20300
The difference in the population of the town at the end of the years 2017 and 2015 was:
(a) 1500 (b) 1700 (c) 1800 (d) 1840
The population of the town at the end of the year 2018 was:
(a) 20100 (b) 20500 (c) 20748 (d) 20850
Which of the following expressions gives the population at the end of the year 2018?
(a) $20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{105}{100}$
(b) $20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{95}{100}$
(c) $20000 \times \frac{104}{100} \times \frac{95}{100} \times \frac{95}{100}$
(d) $20000 \times \frac{96}{100} \times \frac{95}{100} \times \frac{95}{100}$
If the population of the town at the end of 2019 was 21163, then during 2019, the population increases at the rate of:
(a) 2% (b) 3% (c) 3.5% (d) 4%
→1. The population of the town in 2016 was:
(a) 20800 (b) 20500 (c) 20460 (d) 20300
The difference in the population of the town at the end of the years 2017 and 2015 was:
(a) 1500 (b) 1700 (c) 1800 (d) 1840
The population of the town at the end of the year 2018 was:
(a) 20100 (b) 20500 (c) 20748 (d) 20850
Which of the following expressions gives the population at the end of the year 2018?
(a) $20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{105}{100}$
(b) $20000 \times \frac{104}{100} \times \frac{105}{100} \times \frac{95}{100}$
(c) $20000 \times \frac{104}{100} \times \frac{95}{100} \times \frac{95}{100}$
(d) $20000 \times \frac{96}{100} \times \frac{95}{100} \times \frac{95}{100}$
If the population of the town at the end of 2019 was 21163, then during 2019, the population increases at the rate of:
(a) 2% (b) 3% (c) 3.5% (d) 4%
Solve the following equations graphically $:\ 2x + 4y = 7;3x + 8y = 10$
→The cost price of $5$ pens is the same as the selling price of $4$ of them. Find the gain percent.
→Find the greatest and the smallest rational number among the following.$\frac{-2}{3}, \frac{-7}{9}$ and $\frac{-5}{6}$
→A wire when bent in the form of a square encloses an area of $484\ m^2$. Find the largest area enclosed by the same wire when bent to from:$(i)$ An equilateral triangle.$(ii)$ A rectangle of length $16\ m.$
→Find the area and perimeter of the following semicircles: Diameter $= 5.6\ cm$
→