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CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 14 - MCQExams.com
CBSE
Class 10 Maths
Pair Of Linear Equations In Two Variables
Quiz 14
If $$2a = b$$, the pair of equations $$x + 2y = 2a - 6b, ax + by = 2a^{2} - 3b^{2}$$ possess _____ solution(s)
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0%
No
0%
Only one
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Only two
0%
An infinite number of slutions
Explanation
Given $$2a=b\Rightarrow a=\dfrac{b}{2}$$
$$x+2y=2a-6b$$ [first given equation]
$$\Rightarrow x+2y=b-6b=-5b$$ ........eq(i)
$$ax+by=2a^{2}-3b^{2}$$ [second given equation]
$$\Rightarrow \dfrac{b}{2}x+by=\dfrac{2b^{2}}{4}-3b^{2}$$
$$\Rightarrow \dfrac{bx}{2}+by=\dfrac{-5b^{2}}{2}$$
$$\Rightarrow bx+2by=-5b^{2}$$ ..........eq(ii)
From (i) and (ii), Both represents same equations.
So there exits infinite number of solutions.
If the sum of $$2$$ real numbers is $$20$$ and their difference is $$6$$, find the value of their product.
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0%
$$51$$
0%
$$64$$
0%
$$75$$
0%
$$84$$
0%
$$91$$
Explanation
Let the two numbers are $$x$$ and $$y$$
Then according to the question,
$$\Rightarrow x+y=20 ...(1)$$
$$\Rightarrow x-y=6 ...(2)$$
Add $$(1)$$ and $$(2)$$, we get
$$\Rightarrow 2x=26$$
$$\Rightarrow x=\dfrac{26}{2}=13$$
Put the value of $$x$$ in $$(1)$$
$$\Rightarrow 13+y=20$$
$$\Rightarrow y=20-13=7$$
Product of $$x$$ and $$y=$$ $$x\times y=13\times 7=91$$
If the system of linear equations $$\displaystyle\begin{cases}\dfrac12x - \dfrac23y = 7\\ax-8y = -1\end{cases}$$ has no solution then the value of constant $$a$$ is.
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$$-2$$
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$$-\displaystyle\frac{1}{2}$$
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$$2$$
0%
$$6$$
Explanation
If a system of linear equation has no solution then, $$\dfrac{a_{1}}{a_{2}}= \dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}$$
Therefore $$\dfrac{\dfrac{1}{2}}{a}=\dfrac{-\dfrac{2}{3}}{-8}$$
$$\Rightarrow \dfrac{1}{2a}=\dfrac{1}{12}$$
$$\Rightarrow 2a=12$$
Hence $$a=6$$
$$3x+y=19$$, and $$x+3y=1$$. Find the value of $$2x+2y$$.
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$$20$$
0%
$$18$$
0%
$$11$$
0%
$$10$$
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None of the above
$$\displaystyle \begin{cases} \dfrac{3}{4}x-\dfrac{1}{2}y=12\\ \\ kx-2y=22 \end{cases}$$
If the above system of equations has no solution, find the value of $$k$$.
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$$\displaystyle -\frac{4}{3}$$
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$$\displaystyle -\frac{3}{4}$$
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$$3$$
0%
$$4$$
Explanation
Given: $$\dfrac 34 x - \dfrac 12 y -12=0$$ and $$kx-2y-22 = 0$$
$$\therefore a_1 = \dfrac 34, b_1 = -\dfrac 12, c_1 = -12$$
And, $$a_2 = k, b_2 = -2, c_2 = -22$$
Since the given system of equations has no solution,
$$\dfrac {a_1}{a_2} = \dfrac {b_1}{b_2} \neq \dfrac {c_1}{c_2}$$
$$\dfrac {3}{4k} = \dfrac {1}{4} \neq \dfrac {12}{22}$$
Hence, $$k = 3$$
If $$\left| x \right| +x+y=10$$ and $$x+\left| y \right| -y=12$$ then find the value of '$$x+y$$'.
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$$\dfrac { 18 }{ 5 } $$
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$$\dfrac { 17 }{ 5 } $$
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$$\dfrac { 8 }{ 5 } $$
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None of these
Explanation
Assuming $$ y>0 $$ leads to $$ x = 12 $$ from 2nd equation
as: $$ x+y-y = 12 $$
$$ x= 12 $$
and then from 1st Equation $$ |x|+x+y = 10 $$
$$ 2x+y = 10 $$ or $$ y = 14 $$
which is contradiction as $$ y>0 $$
Assuming $$y \leq 0 $$ leads to two cases
(i) if $$ x\leq 0 $$ then from 1st Equation $$ y = 10 $$ and
that contradictory again
(2) if $$ x>0 $$ then $$ 2x+y = 10...(1), x-2y = 12 ...(2) $$
$$(1)\times 2+(2) $$
$$ 4x+2y=20 $$
$$x-2y = 12 $$
__________
$$ 5x = 32 $$
$$ x = \dfrac{32}{5} $$
So on subtituting the value of $$x$$ in equ (2)
$$ \dfrac{32}{5}-2y = 12$$
$$ 2y = \dfrac{32}{5}-12 $$
$$ y = \dfrac{-14}{5} $$
$$ \therefore x+y=\dfrac{32}{5}+\dfrac{-14}{5}$$
$$ \therefore x+y=\dfrac{18}{5}$$
Some students are divided into two groups A & B. If $$10$$ students are sent from A to B, the number in each is the same. But if $$20$$ students are sent from B to A, the number in A is double the number in B. Find the number of students in each group A & B.
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$$100, 80$$
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$$80, 100$$
0%
$$110, 70$$
0%
$$70, 110$$
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None of these
Explanation
Let the number of students in $$A$$ and $$B$$ be $$a \& b$$ respectively.
As per the question:
$$a-10 = b + 10$$
$$a- b = 20 ... (i)$$
and
$$a + 20 = 2 (b -20)$$
$$a - 2b = -60..... (ii)$$
Since coefficient of $$a$$ is same in both equations, we can subract $$(ii)$$ from $$(i)$$.
$$(i)-(ii) :$$
$$a- b -(a-2b) = 20-(-60)$$
$$\implies b =80$$
$$a =80+20 =100$$
Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77 but three tickets from city A to B and two tickets from city A to C cost Rs.What are the fares for cities B and C from A ?
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Rs. $$4$$, Rs. $$23$$
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Rs. $$13$$, Rs. $$17$$
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Rs. $$15$$, Rs. $$14$$
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Rs. $$17$$, Rs. $$13$$
Explanation
Let Rs. $$x$$ be the fare of city B from city A and Rs. $$y$$ be the fare of city C from city A.
Then,
$$2x + 3y = 77 $$...(i)
and
$$3x + 2y = 73$$ ...(ii)
Multiplying (i) by $$3$$ and (ii) by $$2$$ and subtracting,
we get: $$5y = 85$$ or $$y = 17.$$
Putting $$y = 17$$ in (i), we get:$$ x = 13.$$
Henry is three times as old as Truman. Two years ago, Henry was five times as old as Truman. How old is Henry now?
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0%
4
0%
8
0%
12
0%
16
0%
20
A sum of $$Rs. 6.25$$ is made up of $$80$$ coins which are either $$10$$ paise or $$5$$ paise coins. The number of $$10$$ paise and $$5$$ paise coins are _____ and ____ respectively.
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$$45 , 35$$
0%
$$42 , 38$$
0%
$$39, 41$$
0%
$$37 , 43$$
Draw the graph of each of the following linear equations and answer the following question.
(i) y = x (ii) y = 2x (iii) y = -2x (iv) y = 3x (v) y = -3x
these equations are of the form y = mx, where m is a real number.
State True or False : All these graphs pass through origin.
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0%
True
0%
False
Explanation
Yes, these equations $$y=x, y=2x,y=-2x,y=3x$$ and $$y=-3x$$ all pass through origin.
Hence, the statement is true.
Srikant bought two cows for Rs. $$30000$$. By selling one at a loss of $$15\%$$ and the other at a gain of $$19\%$$, he found that selling price of both cows is the same. Find the cost price of each ( in Rs).
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$$10000, 20000$$
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$$15000, 15000$$
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$$17500, 12500$$
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$$17000, 12500$$
Explanation
Let $$x,y$$ be the $$C.P$$ of the $$2$$ cows
$$x+y=3000-1$$ (given)
One is sold at $$15$$% loss and other at gain of $$19$$% S.P of these $$2$$ are equal
$$x-\cfrac { 15 }{ 100 } x=y+\cfrac { 19 }{ 100 } y$$
$$85x=119y\Rightarrow x=\cfrac { 119 }{ 85 } y$$
Using $$1$$ and above result
$$\cfrac { 119 }{ 85 } y+y=30000$$
$$204y=85\times 30000$$
$$y=12,500$$
$$x=30000-12500=17500$$
Cost price of each cow is $$17500,12500$$
The graph of the equations $$x + y = 5$$ and $$x -y = 1$$ will be.....
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Parallel lines
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Intersecting lines
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Lines coincide
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Concurrent lines
Explanation
Take few points on bothe lines and draw the graphs of both lines.
For $$x + y = 5$$:
x
0
3
5
y
5
2
0
For $$x - y = 1$$:
x
0
1
3
y
-1
0
2
From the graphs, we can say that the lines are intesecting at $$(3,2)$$.
Solve the following pairs of equations.
$$4x-y-5=0$$; $$x+y-5=0$$
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$$\left( 2,3 \right) $$
0%
$$\left( 3,3 \right) $$
0%
$$\left( 4,3 \right) $$
0%
None of these
Explanation
$$x+y=5$$ .......eq1;
$$4x-y=5$$ ......eq2
Add both the equations, we will get $$x=2$$
Put $$x= 2$$ in eq1, we get
$$y=3$$
Solve the following pairs of equations.
$$2x-4=0$$; $$4x+y+4=0$$
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$$\left( 2,-12 \right) $$
0%
$$\left( 1,-12 \right) $$
0%
$$\left( 2,12 \right) $$
0%
None of these
Explanation
$$2x-4= 0 \implies x=2$$
and
$$4x+y =-4$$
$$4(2)+y=-4$$
$$8+y =-4$$
$$y =-12$$
$$\Rightarrow x=2, y = -12 $$
Solve the following pairs of equations.
$$3x-y=0$$; $$x-2=0$$
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$$x=2,y=6$$
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$$x=3,y=6$$
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$$x=4,y=6$$
0%
None of these
Explanation
As we are give $$3x-y=0$$ & $$x-2= 0 $$
$$\therefore$$ $$x=2$$ &, on substituting the value of $$x$$ in $$3x-y=0$$
$$\therefore y=6$$
Solve the following pairs of equations.
$$x-y=0$$; $$y+3=0$$
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$$\left( -3,-3 \right) $$
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$$\left( 3,-3 \right) $$
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$$\left( 2,-3 \right) $$
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None of these
Explanation
Consider $$x-y=0$$
$$\Rightarrow x=y$$
$$y+3=0 \implies y =-3$$
$$\Rightarrow x= y =-3$$
Hence solution for given equations is $$(-3,-3)$$
Solve the following pairs of equations.
$$2x=y+1$$; $$x+2y-8=0$$
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$$x= 2,y=3 $$
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$$x= 3,y=3 $$
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$$x= 2,y=2 $$
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None of these
Explanation
$$2x=y+1$$.....(ii) ;
and
$$x+2y=8$$
$$x=8-2y$$.....(i)
Put (i) in (ii)
$$16-4y=y+1$$
$$y=3$$
$$y=3$$ & $$x=2$$
Solve the following pairs of equations.
$$x+y=5$$; $$x-y=1$$
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$$\left( 3,2 \right) $$
0%
$$\left( 2,2 \right) $$
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$$\left( 4,2 \right) $$
0%
None of these
Explanation
$$x-y = 1$$ ....eq1
$$x+y=5$$.....eq2
Add & subtract both the equations, we will get
$$2X=6$$
$$\implies x=3$$
&
Put the value of x in equ 1
$$y=2$$
Solve the following pairs of equations.
$$y=2x+1$$; $$y+3x-6=0$$
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$$\left( 1,3 \right) $$
0%
$$\left( 2,3 \right) $$
0%
$$\left( 4,3 \right) $$
0%
None of these
Explanation
Here given equations are $$y-2x= 1$$ ; $$y+3x=6$$
Subtract both the equations, we will get $$x=1$$.
On substituting the value of $$x$$ in any given equations we get
$$y=3$$
Solve the following equations by substitution method.
$$x + 3y = 10; 2x + y = 5$$
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$$x = 1, y = 3$$
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$$x = 2, y = 3$$
0%
$$x = 1, y = 1$$
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None of these
Explanation
$$x+3y=10$$ .....$$(1)$$ ;
$$2x+y=5$$........$$(2)$$
$$y=5-2x$$......from $$(2)$$
putting in $$(1)$$, we get
$$\Rightarrow x+3(5-2x)=10$$
$$\Rightarrow x+15-6x=10$$
$$\Rightarrow -5x=-5$$
$$\Rightarrow x=1$$
$$1+3y=10$$
$$y=3$$
Solve the system of equations
$$65x-33y=97, $$ and $$33x-65y=1$$
by
substitution
method.
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$$(2, 1)$$
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$$(1, 1)$$
0%
$$(2, 2)$$
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None of these
Explanation
Solve the following equations by substitution method.
$$\dfrac{1}{x} + \dfrac{2}{y} = 9; \dfrac{2}{x} + \dfrac{1}{y} = 12 (x \neq 0, y \neq 0)$$
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$$x = \dfrac{1}{5} , y = \dfrac{1}{2}$$
0%
$$x = \dfrac{2}{5} , y = \dfrac{1}{2}$$
0%
$$x = \dfrac{1}{5} , y = \dfrac{-1}{2}$$
0%
None of these
Explanation
Given equations,
$$\dfrac{1}{x} + \dfrac{2}{y} = 9$$..............(1)
$$\dfrac{2}{x} + \dfrac{1}{y} = 12$$..............(2)
$$\dfrac{1}{x}=9-\dfrac{2}{y}$$.........from (1)
putting value of $$\dfrac {1}{x}$$ in (2), we get
$$2\left[9-\dfrac{2}{y}\right]+\dfrac{1}{y}=12$$
$$18-\dfrac{3}{y}=12$$
$$6=\dfrac{3}{y}$$......
$$\boxed{y=\dfrac{1}{2}}$$
putting value of $$y$$ in (1), we get
$$\dfrac {1}{x}=9-4$$
$$\dfrac{1}{x}=5$$......$$\boxed{x=\dfrac{1}{5}}$$
Solve the following equations by substitution method.
$$\dfrac{3}{x} + \dfrac{1}{y} = 7; \dfrac{5}{x} - \dfrac{4}{y} =(x \neq 0, y \neq 0)$$
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$$x = \dfrac{1}{2}, y = 1$$
0%
$$x = \dfrac{-1}{2}, y = 1$$
0%
$$x = \dfrac{1}{2}, y = -1$$
0%
None of these
Explanation
Given equations are,
$$\dfrac{3}{x}+\dfrac{1}{y}=7$$
$$\dfrac{5}{x}-\dfrac{4}{y}=6$$
$$\dfrac{1}{x}=\dfrac{1}{3}\left(7-\dfrac{1}{y}\right)$$.................from (1)
Putting value of $$\dfrac {1}{x}$$ in (1), we get
$$5\left[\dfrac{1}{3}(7-\dfrac{1}{y})\right]-\dfrac{4}{y}=6$$
$$\dfrac{35}{3}-\dfrac{5}{3y}-\dfrac{4}{y}=6$$
$$\dfrac{17}{3}=\dfrac{5+12}{3y}$$
$$\boxed {y=1}$$
Putting value of $$y$$ in (1)
$$\dfrac{1}{x}=\dfrac{1}{3}\times 6\Rightarrow \dfrac {1}{x}=2=\boxed{x=\dfrac{1}{2}}$$
Solve following system of equations by elimination method. $$x +2y=7, x-2y=1$$
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0%
$$\left ( 4, \dfrac{3}{2} \right )$$
0%
$$\left ( 4, \dfrac{2}{2} \right )$$
0%
$$\left ( 3, \dfrac{3}{2} \right )$$
0%
None of these
Explanation
Given equations are
$$x+2y=7$$.......(i)
$$x-2y=1$$.........(ii)
Subtract equation (ii) from (i), we get
$$2x=8$$
$$\Rightarrow x=4$$
put value of $$x$$ in equation (ii)
$$4-1=2y$$
$$3=2y$$
$$\therefore y=\dfrac{3}{2}$$
$$\therefore x=4 , \ y=\dfrac{3}{2}$$
Solve each of the following system of equations by substitution method.
$$13x +11y=70, 11x+13y=74$$
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0%
$$(2, 4)$$
0%
$$(3, 4)$$
0%
$$(2, 2)$$
0%
None of these
Explanation
Since, $$13x+11y=70$$...eq1 $$11x+13y=74$$......eq2
$$x=\dfrac{70-11y}{13}$$, put this value in eq2, we get
$$11\left[\dfrac{70-11y}{13}\right]+13y=74$$
$$770-121y+169y=74\times 13$$
$$\Rightarrow y=4$$
and
$$x=\dfrac{70-11(4)}{13}$$
$$x=2$$
Solve the following equations by substitution method.
$$5x + 3y = 21; 2x - y = 4$$
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0%
$$x = 3, y = 2$$
0%
$$x =- 3, y = 2$$
0%
$$x = 3, y = -2$$
0%
None of these
Explanation
$$5x+3y=21$$.........(1)
$$2x-y=4$$...............(2)
$$y=2x-4$$
$$5x+3(2x-4)=21$$.........putting value of $$y$$ in (1)
$$\Rightarrow 5x+6x-12=21$$
$$\Rightarrow 11x=33$$.......................$$\boxed{x=3}$$
Putting value of $$x$$ in (2)
$$6-y=4$$..........................
$$\boxed{y=2}$$
Solve the following equations by substitution method.
$$2x + y = 1; 3x - 4y = 18$$
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0%
$$x = 2, y = - 3$$
0%
$$x = -2, y = - 3$$
0%
$$x = 2, y = 3$$
0%
None of these
Explanation
$$2x+y=1$$...........(1)
$$3x-4y=18$$........(2)
$$y=1-2x$$...........from (1)
putting value of $$y$$ in (1)
$$\Rightarrow 3x-4(1-2x)=18$$
$$\Rightarrow 3x-4+8x=18$$
$$\Rightarrow 11x=22$$
$$\Rightarrow x=2$$
Using value of $$x$$
$$\Rightarrow 4+y=1$$
$$\Rightarrow y=-3$$
Solve the following system of equations by substitution method.
$$\dfrac{15}{x}+\dfrac{2}{y}=17, \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}, x\neq 0, y\neq 0$$
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0%
$$\left ( 5, \dfrac{1}{7} \right )$$
0%
$$\left ( 2, \dfrac{1}{7} \right )$$
0%
$$\left ( 5, \dfrac{2}{7} \right )$$
0%
None of these
Explanation
Given equations are
$$\dfrac{15}{x}+\dfrac{2}{y}=17,$$
$$ \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5} $$
$$Let \ \dfrac{1}{x}=a, \ \ \dfrac{1}{y}=b$$
$$\therefore \ 15a+2b=17$$......(i)
and $$a+b=\dfrac{36}{5}$$
$$\Rightarrow a=\dfrac{36}{5}-b$$....(ii)
Putting equation (ii) in (i) we get,
$$15\left[\dfrac{36}{5}-b\right]+2b=17$$
$$\Rightarrow 108-13b=17$$
$$\Rightarrow b=7$$
$$\therefore \ y=\dfrac{1}{7}$$
Putting value of $$b$$ in equation (i), we get,
$$15a+(2\times7)=17$$
$$\Rightarrow a=\dfrac{1}{5}$$
$$\therefore \ x=5$$
Hence, $$\left(5,\dfrac17\right)$$ is the solution of the given system of equations.
The sum of two numbers is $$25$$ and their difference is $$7$$, then the numbers are.
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0%
$$20$$ and $$5$$
0%
$$18$$ and $$7$$
0%
$$15$$ and $$10$$
0%
$$9$$ and $$16$$
Explanation
let the two numbers be$$ A$$ and $$B$$
so it is given that
$$ A + B = 25 $$...(1)
$$ A - B = 7 $$ ...(2)
Adding the equations we get
$$ A + B + A - B = 25 +7 $$
$$ 2A = 32 $$
$$A = 16$$
Using the above value of A in (1)
So $$ 16 + B = 25 $$
$$=>B = 9$$
Thus $$ A =16$$ and $$B =9 $$
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