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CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 12 - MCQExams.com
CBSE
Class 10 Maths
Pair Of Linear Equations In Two Variables
Quiz 12
Solve graphically: $$5x-6y+30=0$$; $$5x+4y-20=0$$. Also, find the vertices of the triangle formed by the above two lines and $$x$$-axis.
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$$(3,5),(4,2),(6,9)$$
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$$(0,5),(-6,0),(4,0)$$
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$$(0,2),(7,2),(0,4)$$
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$$(2,3),(1,7),(3,9)$$
Explanation
For $$5x - 6y + 30 = 0$$:
x
0
-6
6
y
5
0
10
For $$x - y - 1 = 0$$:
x
0
4
-4
y
5
0
10
As the graph shows, vertices of triangle are $$(0, 5), (-6, 0) \& (4, 0)$$.
In a zoo there are some pigeons and some rabbits. If the total number of heads is $$300$$ and the total number of legs is $$750$$, then how many pigeons are there?
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$$150$$
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$$175$$
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$$200$$
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$$225$$
Explanation
Let, the total number of pigeons be $$p$$ and the total number of rabbits be $$r$$
Both the pigeons and rabbits has one head each
so, $$p+r=300$$ ...........................................(1)
Each pigeon has two legs and each rabbit has four legs
so, $$2p+4r=750$$................................................(2)
Multiply by (1) by $$4$$
Then $$4p+ 4r=1200$$................................................(3)
By subtraction (2) from (3)
$$(4p+ 4r)-(2p+4r)=1200-750$$
$$\Rightarrow 4p+ 4r-2p-4r=450$$
$$\Rightarrow 2p=450$$
$$\Rightarrow p=225$$
Hence, the number of pigeons in the zoo is $$225$$.
Using substitution method find the value of x and y:
$$2x - 3y = 2$$ and $$5x - 6y = 2$$
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-4 and -1
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-2 and 10
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-2 and -11
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-2 and -2
Explanation
$$2x- 3y = 2$$ ------- (1)
$$5x- 6y = 2$$------ (2)
From equation 1:
$$2x- 2 = 3y$$
$$y =\dfrac{2x-2}{3}$$
Substitute the value of y in equation 2:
$$5x- 6y = 2$$
$$5x- 6(\dfrac{2x-2}{3}) = 2$$
$$5x- 2(2x- 2) = 2$$
$$5x- 4x + 4 = 2$$
$$x = 2- 4$$
$$x = -2$$
Now, Substitute $$x = -2$$ in equation 1:
$$5x- 6y = 2$$
$$5(-2)- 6y = 2$$
$$-10 -2 = 6y$$
$$6y = -12$$
$$y = \dfrac{-12}{6} = -2$$
Therefore the solution is: $$x = -2$$ and $$y = -2$$
The total salary of $$15$$ men and $$8$$ women is $$Rs.\ 3050$$. The difference of salaries of $$5$$ women and $$3$$ men is $$Rs.\ 50$$. Find the sum of the salaries of $$3$$ men and $$3$$ women
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$$Rs.\ 900$$
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$$Rs.\ 850$$
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$$Rs.\ 750$$
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$$Rs.\ 1000$$
Explanation
Let the salary of $$1$$ men be $$x$$ and the salary of $$1$$ woman be $$y.$$
As per the question,
$$15x + 8y =3050\quad\quad...(1)$$
$$3x-5y=-50\quad\quad...(2)$$
Multiply equation $$(2)$$ by $$5$$,
$$15x -25y = -250\quad\quad...(3)$$
Subtract equation $$(3)$$ from equation $$(1),$$
$$\left( {15x + 8y} \right) - \left( {15x - 25y} \right) = 3050 - \left( { - 250} \right)$$
$$\Rightarrow 15x + 8y - 15x + 25y = 3050 + 250$$
$$\Rightarrow 33y = 3300$$
$$\Rightarrow y= 100$$
Substitute the value of $$y$$ in equation $$(2),$$
$$3x-500 =-50$$
$$\Rightarrow 3x= 450$$
$$\Rightarrow x= 150$$
Now, total salary of $$3$$ men and $$3$$ women will be,
$$3x + 3y= 3 \times 150 + 3 \times 100$$
$$=450 + 300$$
$$ =Rs.\ 750$$
A students walks from his house at less 4 km per hour and reaches his school late by 5 minutes. If his speed has been increase by 5 km per hour then he would have reached 10 minutes early. The distance of the school from his house is
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$$ \displaystyle \frac{3}{5}km $$
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$$5$$km
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$$6$$km
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$$4$$km
Explanation
Let the actual time be 't'
and distance be x
Now according to first condition, we have
$$\dfrac{x}{4}=t+\dfrac{5}{60}$$
And according to Second condition, we have
$$\dfrac{x}{5}=t-\dfrac{10}{60}$$
$$60x=4t+20$$
$$60x=5t-50$$
Subtracting equation 1 from 2, we have
$$0=t-70$$
$$\therefore t=70$$
Putting the value of t we get $$x=5km$$
Solve for $$x$$ and $$y$$, if $$2x+ 3y= 8$$ and
$$x+2y=5$$.
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$$x=1, y=2$$
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$$x=-1, y=-2$$
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$$x=-2, y=2$$
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$$x=-1, y=2$$
Explanation
$$2x+3y=8$$ .............(1)
and $$x+2y=5$$ .......(2)
From (1), we get $$x=\frac {8-3y}{2}$$ .......... (3)
From (2), we get $$x=5-2y$$ .......... (4)
$$\therefore \frac {8x-3y}{2}=5-2y$$
or $$8x -3y= 10 -4y$$ or $$3y= 10 -4y$$ or $$y= 2$$
Using $$y = 2$$ in (4), we get $$x = 5 -4 = 1$$. Therefore,
$$x=1,y=2$$
The sum of the digits of a two-digit number is $$5$$. The digit obtained by increasing the digit in tens' place by unity is one-eighth of the number. Then the number is
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less than $$30$$
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lies between $$30$$ and $$40$$
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more than $$37$$
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lies between $$40$$ and $$50$$
Explanation
$$x+y=5$$
and $$x+1=\dfrac {1}{8}(10x+y)$$ .........(1)
or $$8x+8=10x+y$$
or $$2x+y=8$$ .........(2)
Subtract equation (1) from equation (2)
$$2x+y=8$$
$$+ x+y=5$$
$$\underline {(-) (-) (-)}$$
$$x=3$$
$$\therefore y=2$$
Number is $$32$$.
10 years ago the age of the father was 5 times that of the son. 20 years hence the age of the father will be twice that of the son. The present age of the father (in years) is
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$$40$$
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$$45$$
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$$60$$
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$$70$$
Explanation
Let the present age of father be $$x$$ years. and that of son be $$y$$ years
10 years ago the age of the father was 5 times that of the son.
$$\therefore$$ $$x-10=5(y-10)$$
$$\therefore$$ $$x-5y=-40$$ ...(1)
20 years hence, the age of the father will be twice that of the son.
$$\therefore$$ $$x+20=2(y+20)$$
$$\therefore$$ $$x-2y=20$$ ...(2)
Subtracting (1) and (2) we get ,
$$y=20$$ and $$x=60$$
$$\therefore$$ The present age of father is $$60$$ years
The graphs of $$2x+3y-6=0, 4x-3y-6=0, x=2$$ and $$ \displaystyle y=\dfrac{2}{3} $$ intersect in
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four points
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one point
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in no points
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infinite number of points
Explanation
Solving $$2x + 3y - 6 = 0$$ and $$4x - 3y - 6 = 0$$ we get
$$x = 2$$ and $$\displaystyle y=\dfrac{2}{3}$$
Hence from all the four given equations the
value of $$x$$ is $$2$$ and $$y$$ is $$\displaystyle \frac{2}{3}$$
Therefore the graphs of all the four lines intersect
in one point
What values of x and y satisfies the equations $$\displaystyle \frac{x}{6}+\frac{y}{15}=4$$ and $$\displaystyle \frac{x}{3}-\frac{y}{12}=4\frac{3}{4}$$
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$$(6,15)$$
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$$(18,15)$$
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$$(15,18)$$
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$$(12,30)$$
Explanation
We pick either of the equations and write one variable in terms of the other.
Let us consider the equation
$$\frac { x }{ 3 } -\frac { y }{ 12 } =4\frac { 3 }{ 4 }$$
or
$$4x-y=57$$
and write it as
$$y=4x-57...........(1)$$
Substitute the value of $$y$$ in Equation
$$\frac { x }{ 6 } +\frac { y }{ 15 } =4$$ or
$$15x+6y=360$$
. We get
$$15x+6(4x-57)=360$$
i.e.
$$15x+24x-342=360$$
i.e.
$$39x=702$$
i.e.
$$x=18$$
Therefore,
$$x=18$$
Substituting this value of $$x$$ in
equation
1
, we get
$$y=72-57$$
i.e.
$$y=15$$
Hence, the solution is $$x = 18, y =15$$.
The ratio of the ages of A and B ten years before was $$3 : 5$$. The ratio of their present ages is $$2 : 3$$ . Their ages are respectively
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$$30,50$$
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$$20,30$$
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$$40,60$$
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$$16,24$$
Explanation
Let the present ages of $$A$$ and $$B$$ are $$x$$ and $$y$$ years respectively
$$\displaystyle \therefore \dfrac{x-10}{y-10}=\dfrac{3}{5}$$ .....(i)
or $$5x -50=3y-30$$ ......(ii)
and $$\displaystyle \dfrac{x}{y}=\dfrac{2}{3} $$or $$3x=2y$$ or $$y=\dfrac{3x}{2}$$
putting the value of y in equation (i) we get
$$5x-50=$$$$\displaystyle \dfrac{9x}{2}-30$$
$$10x-100=9x-60$$
or $$x=40$$ $$\displaystyle \therefore y=\dfrac{3\times 40}{2}=60$$
Solve the equations using graphical method: $$x + y = 1$$ and $$2x -3y = 7$$
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(2, -1)
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(-2, -1)
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(-2, 1)
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(2, 1)
Explanation
$$x + y = 1$$ ------- (I)
$$2x - 3y = 7$$ ------ (II)
Find the value of x and y using substitution method:
$$5x - 6y = 2$$ and $$6x - 5y = 9$$
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$$4$$ and $$-3$$
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$$-3$$ and $$-4$$
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$$-4$$ and $$-3$$
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$$4$$ and $$3$$
Explanation
$$5x -6y = 2$$ ------- $$(1)$$
$$6x - 5y = 9$$ ------- $$(2)$$
From equation $$(1)$$:
$$\Rightarrow 6y = 5x- 2$$
$$\Rightarrow y =\dfrac{5x-2}{6}$$
Substitute the value of $$y$$ in equation $$(2)$$ :
$$\Rightarrow 6x - 5y = 9$$
$$\Rightarrow6x- 5(\dfrac{5x-2}{6}) = 9$$
$$\Rightarrow36x -25x + 10 = 54$$
$$\Rightarrow11x = 54- 10$$
$$\Rightarrow x = \dfrac{44}{11} = 4$$
Now, Substitute $$x = 4$$ in equation $$(1)$$
$$\Rightarrow5x - 6y = 2$$
$$\Rightarrow5(4)-6y = 2$$
$$\Rightarrow 20 - 2 = 6y$$
$$\Rightarrow18 = 6y$$
$$\Rightarrow y =\dfrac{18}{6}=3$$
$$\therefore$$ the solution is $$x = 4$$ and $$y = 3$$
Find the value of $$x$$ and $$y$$ using substitution method:
$$x- y = 2$$ and $$2x - y = 9$$
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$$7$$ and $$7$$
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$$7$$ and $$5$$
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$$-7$$ and $$-5$$
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$$5$$ and $$7$$
Explanation
$$x- y = 2$$ ------- $$(1)$$
$$2x - y = 9$$ ------ $$(2)$$
From equation $$(1)$$ :
$$y = x - 2$$
Substitute the value of $$y$$ in equation $$(2)$$:
$$2x - y = 9$$
$$2x -x + 2 = 9$$
$$x = 9 -2$$
$$x = 7$$
Now, Substitute $$x = 7$$ in equation $$(1)$$:
$$x- y = 2$$
$$7- y = 2$$
$$y = 7 -2$$
$$y = 5$$
Therefore the solution is: $$x = 7$$ and $$y = 5$$
Solve equations using substitution method: $$x-y = 3$$ and $$x + y = 0$$
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$$\dfrac{3}{2}$$ and $$1$$
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$$\dfrac{-3}{2}$$ and $$-1$$
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$$\dfrac{3}{2}$$ and $$-\dfrac{3}{2}$$
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$$0$$ and $$\dfrac{3}{2}$$
Explanation
$$x- y = 3\dots (1)$$
$$x + y = 0\dots (2)$$
From equation $$(1)$$ :
$$y = x - 3$$
Substitute the value of $$y$$ in equation $$(2)$$ :
$$x + y = 0$$
$$x + x- 3 = 0$$
$$2x = 3$$
$$x=\dfrac{3}{2}$$
Now, Substitute $$x = \dfrac{3}{2}$$ in equation $$(1)$$:
$$x- y = 3$$
$$\dfrac{3}{2}- y = 3$$
$$y = \dfrac{3}{2}- 3$$
$$y = -\dfrac{3}{2}$$
Therefore the solution is:
$$x =\dfrac{3}{2}$$ and $$y = -\dfrac{3}{2}$$
Solve equations using substitution method:
$$2x + y = 5$$ and $$2x- 3y = 3$$
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$$\frac{1}{4}$$ and $$\frac{-1}{2}$$
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$$\frac{1}{4}$$ and $$\frac{1}{2}$$
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$$\frac{-9}{4}$$ and $$\frac{-1}{2}$$
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$$\frac{9}{4}$$ and $$\frac{1}{2}$$
Explanation
$$2x + y = 5$$ ------- (1)
$$2x- 3y = 3$$------ (2)
From equation 1 :
$$y = 5 -2x$$
Substitute the value of y in equation 2 :
$$2x -3(5- 2x) = 3$$
$$2x -15 + 6x = 3$$
$$8x = 3 + 15$$
$$x = \dfrac{18}{8} = \dfrac{9}{4}$$
Now, Substitute $$x = \dfrac{9}{4}$$ in equation 1 :
$$2(\dfrac{9}{4}) + y = 5$$
$$18 + 4y = 20$$
$$4y = 20 -18$$
$$y = \dfrac{2}{4}$$
$$y = \dfrac{1}{2}$$
Therefore the solution is: $$x = \dfrac{9}{4}$$ and $$y = \dfrac{1}{2}$$
Solve equations using substitution method:
$$2x -y = 3$$ and $$4x + y = 3$$
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$$-1$$ and $$1$$
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$$-2$$ and $$1$$
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$$1$$ and $$-1$$
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$$-1$$ and $$-1$$
Explanation
$$2x- y = 3$$ ------- (1)
$$4x + y = 3$$ ------ (2)
From equation 1 :
$$y = 2x - 3$$
Substitute the value of $$y$$ in equation 2 :
$$4x + y = 3$$
$$4x + 2x -3 = 3$$
$$6x = 3 + 3$$
$$x = \dfrac{6}{6} = 1$$
Now, Substitute $$x = 1$$ in equation 1 :
$$2x -y = 3$$
$$2 (1)- y = 3$$
$$2- 3 = y$$
$$y = -1$$
Therefore the solution is: $$x = 1$$ and $$y = -1$$
If $$2$$ tables and $$3$$ chairs cost $$Rs. 4500$$ and $$3$$ tables and $$2$$ chairs cost $$Rs. 5000$$, then how much does a table cost?
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Rs. $$1100$$
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Rs. $$1200$$
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Rs. $$1150$$
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Rs. $$1300$$
Explanation
Let the cost of table and chair is $$x$$ and $$y$$.
Then, $$2x+3y=4500$$ ....(1)
and $$3x+2y=5000$$ ...(2)
Add eq(1) and eq(2)
$$5x+5y=9500$$
$$x+y = 1900$$ ....(3)
Subtract eq(1) from eq(2)
$$x-y= 500 $$ ....(4)
Solving eq(3) and eq(4) we get
$$2x=2400$$
$$x =1200$$
Father says to son "I am $$5$$ times as old as you were when i was as old as you are". If sum of their ages is $$48$$, what is the age of the father?
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$$20$$
0%
$$9$$
0%
$$45$$
0%
$$40$$
Explanation
Let the present age of father and son is x and y
Then as per question $$x+y=48$$ ........................(1)
Given $$x=5y$$ .......................................................(2)
substituting equation 2 in equation 1 we get
$$5y+y=48$$
$$6y=48$$
$$y=8$$.........................................................................(3)
Putting value of y in eqn 2 we get $$x=5(8)=40$$
We get $$x=40$$
Then age of father is $$40$$ years
Graph the equation by determining the missing values needed to plot the ordered pairs.
$$y+x=4;$$
$$\left( 1, y_1\right) ,\left( 4, y_2 \right) ,\left( 3, y_3\right) $$
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0%
0%
0%
Explanation
We have given line, $$y+x=4\Rightarrow y=4-x$$
Now, at $$x=1, y=4-1=3$$
At $$x=4, y=4-4=0$$ and at $$x=3, y=4-3=1$$
Hence, given points on the line are $$(1,3),(4,0),(3,1)$$.
Clearly these points lying on the graph of line in option B.
Hence option B is correct choice.
Use the method of substitution to solve the equations
$$x+2y=-4$$ and $$4x+5y=2$$
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$$\dfrac{-16}{3}$$ and $$\dfrac{-2}{3}$$
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$$\dfrac{-16}{3}$$ and $$\dfrac{2}{3}$$
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$$8$$ and $$-6$$
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$$\dfrac{16}{3}$$ and $$\dfrac{2}{3}$$
Explanation
$$x+2y=-4$$
$$\implies x=-4-2y$$
Substituting
$$ x=-4-2y$$ in
$$4x+5y=2$$
$$\implies 4(-4-2y)+5y=2$$
$$\implies -16-8y+5y=2$$
$$\implies -16-3y=2$$
$$\implies -16-2=3y$$
$$\implies y=\dfrac{-18}{3}=-6$$
Substituting $$y=-6$$ in
$$ x=-4-2y$$, we get,
$$x=-4-2(-6)$$
$$=-4+12=8$$
$$\therefore x=8, y=-6$$
Solve the equation using substitution method:
$$3x-5y =11$$ and $$+10y+6x=7$$
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$$\left(\frac{29}{12}, \frac{3}{4}\right)$$
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$$\left(\frac{3}{4}, \frac{29}{12}\right)$$
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$$\left(\frac{-29}{12}, \frac{3}{4}\right)$$
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$$\left(\frac{29}{12}, \frac{-3}{4}\right)$$
Explanation
$$3x-5y =11$$ ---------- (1)
$$+10y+6x=7$$ ---------(2)
From equation (1)
$$3x-5y=11$$
$$\frac{3x-11}{5}=y$$
Substitute the value of $$y$$ in equation (2)
$$+10\left(\frac{3x-11}{5}\right)+6x = 7$$
$$+6x-22+6x = 7$$
$$12x=7+22$$
$$x=\frac{29}{12}$$
Substitute the value of $$x$$ in equation (1)
$$3x-5y = 11$$
$$3\left(\frac{29}{12}\right) - 5y = 11$$
$$29-20y = 44$$
$$29-44=20y$$
$$\frac{-153}{20} = y$$
$$y=\frac{-3}{4}$$
therefore solution is $$x=\frac{29}{12}$$ and $$y=\frac{-3}{4}$$
Using substitution method find the value of $$x$$ and $$y:$$
$$x + 4y = -4$$ and $$2x- 3y = 2$$
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$$\dfrac{-4}{11}$$ and $$\dfrac{-10}{11}$$
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$$\dfrac{-2}{11}$$ and $$\dfrac{10}{11}$$
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$$\dfrac{-2}{11}$$ and $$\dfrac{-10}{11}$$
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$$\dfrac{2}{11}$$ and $$\dfrac{-2}{11}$$
Explanation
$$x + 4y = -4$$ ------- $$(1)$$
$$2x- 3y = 2$$ ------ $$(2)$$
From equation $$1:$$
$$4y = -4 -x$$
$$y =\dfrac{-4-x}{4}$$
Substitute the value of $$y$$ in equation $$2:$$
$$2x- 3y = 2$$
$$2x- 3\left(\dfrac{-4-x}{4}\right) = 2$$
$$8x + 12 + 3x = 8$$
$$11x = 8-12$$
$$11x = -4$$
$$x = \dfrac{-4}{11}$$
Now, Substitute $$x =\dfrac{-4}{11}$$ in equation $$1:$$
$$x + 4y = -4$$
$$\dfrac{-4}{11} + 4y = -4$$
$$-4 + 44y = -44$$
$$44y = -44 + 4$$
$$44y = -40$$
$$y = \dfrac{-40}{44} \\= \dfrac{-10}{11}$$
Therefore the solution is: $$x = \dfrac{-4}{11}$$ and $$y = \dfrac{-10}{11}$$
Using substitution method find the value of x and y:
$$3x+5y$$ $$= 1$$ and $$4x + 5y $$$$= 2$$
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$$-1$$ and $$\dfrac{-2}{5}$$
0%
$$-1$$ and $$\dfrac{2}{5}$$
0%
$$1$$ and $$\dfrac{-2}{5}$$
0%
$$1$$ and $$\dfrac{2}{5}$$
Explanation
$$3x+5y=1$$
$$\implies 3x=1-5y$$
$$\implies x=\dfrac{1-5y}{3}$$
Now, Substituting $$x=\dfrac{1-5y}{3}$$
in
$$4x+5y=2$$
$$\implies 4(\dfrac{1-5y}{3})+5y=2$$
$$\implies \dfrac{4-20y+15y}{3}=2$$
$$\implies 4-5y=6$$
$$\implies -2=5y$$
$$\implies y=\dfrac{-2}{5}$$
Substituting $$y=\dfrac{-2}{5}$$ in $$x=\dfrac{1-5y}{3}$$
$$x=\dfrac{1-5(\dfrac{-2}{5})}{3}$$
$$x=\dfrac{1+2}{3}=1$$
$$\therefore x=1, y=\dfrac{-2}{5}$$
Using substitution method find the value of x and y:
$$4x + 9y = 5$$ and $$-5x + 3y = 8$$
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$$-1$$ and $$\dfrac{-2}{5}$$
0%
$$-1$$ and $$\dfrac{2}{5}$$
0%
$$-1$$ and $$1$$
0%
$$1$$ and$$\dfrac{2}{5}$$
Explanation
$$4x+9y=5$$
$$\implies 4x=5-9y$$
$$\implies x=\dfrac{5-9y}{4}$$
Substituting $$x=\dfrac{5-9y}{4}$$ in
$$-5x+3y=8$$, we get,
$$-5(\dfrac{5-9y}{4})+3y=8$$
$$\implies \dfrac{-25+45y+12y}{4}=8$$
$$\implies -25+45y+12y=8\times 4$$
$$\implies -25+57y=32$$
$$\implies 57y=57$$
$$\implies y=1$$
Substituting $$y=1$$in $$x=\dfrac{5-9y}{4}$$
$$x=\dfrac{5-9(1)}{4}$$
$$\implies x=\dfrac{-4}{4}$$
$$\implies x=-1$$
$$\therefore x=-1, y=1$$
Solve the equation using substitution method:
$$3x+y+1=0$$ and $$3x-4y=10$$
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(0.4, 2.2)
0%
(0.4, -2.2)
0%
(0.4, 1.2)
0%
(-0.4, 2.2)
Explanation
$$3x+y+1=0$$ ------- (1)
$$3x-4y=10$$ --------(2)
From equation (1)
$$3x+y=-1$$
$$y=-1-3x$$
Substitute the value of $$y$$ in equation (2)
$$3x-4(-1-3x)=10$$
$$3x+4+12x=10$$
$$15x=10-4$$
$$x=\dfrac{6}{15}$$ or $$\dfrac{2}{5}\simeq 0.4$$
The substitute the value of x in equation (1)
$$3x+y=-1$$
$$3(\dfrac{2}{5})+y=-1$$
$$6+5y=-5$$
$$5y=-5-6$$
$$5y=-11$$
$$y=\dfrac{-11}{5}$$ or -2.2
therefore solution is $$x= 0.4$$ and $$y=-2.2$$
Using substitution method find the value of x and y:
$$x - 9y = 18$$ and $$-x + 3y = -15$$
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$$-13.5 $$ and $$-3.5$$
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$$-13.5$$ and $$2.5$$
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$$13.5$$ and $$-2.5$$
0%
$$13.5$$ and $$-0.5$$
Explanation
$$x-9y=18$$
$$\implies x=18+9y$$
Substituting
$$ x=18+9y$$ in
$$-x+3y=-15$$, we get,
$$ -(18+9y)+3y=-15$$
$$\implies -18-9y+3y=-15$$
$$\implies -6y=-15+18$$
$$\implies -6y=3$$
$$\implies y=-0.5$$
Substituting $$y=-0.5$$ in
$$ x=18+9y$$
$$x=18+9(-0.5)$$
$$\implies x=18-4.5$$
$$\implies x=13.5$$
$$\therefore x=13.5, y=-0.5$$
Solve the equations using the method of substitution method :
$$3x+4y=-43$$ and $$-2x+3y=11$$
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7 and 5
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5 and -7
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-5 and 7
0%
5 and 7
Explanation
$$3x+4y=-43$$ ------- (1)
$$-2x+3y=11$$ ------ (2)
From equation 1:
$$3x+4y=-43$$
$$4y=43-3x$$
$$y=\frac{43-3x}{4}$$
Substitute the value of $$y$$ in equation 2:
$$-2x+3(\frac{43-3x}{4})=11$$
$$-8x + 129-9x= 44$$
$$-17x = 44-129$$
$$x =\frac{-85}{-17}=5$$
The substitute the value of $$x$$ in equation 1:
$$3x+4y=43$$
$$3(5)+4y=43$$
$$15+4y=43$$
$$4y=43-15$$
$$4y=28$$
$$y=\frac{28}{4}$$
$$y=7$$
Therefore the solution is:
$$x = 5$$ and $$y=7$$
Find the value of a and b using substitution method:
$$a + b = 3$$ and $$-3a + 2b = 1$$
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1 and 2
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2 and 1
0%
-2 and -1
0%
2 and -1
Explanation
$$a + b = 3$$ ------- (1)
$$-3a + 2b = 1$$ ------ (2)
From equation 1:
$$a = 3 - b$$
Substitute the value of a in equation 2 :
$$-3a + 2b = 1$$
$$-3(3- b) + 2b = 1$$
$$-9 + 3b + 2b = 1$$
$$5b = 10$$
$$b = \dfrac{10}{5} = 2$$
Now, Substitute $$b = 2$$ in equation 1:
$$a + b = 3$$
$$a + 2 = 3$$
$$a = 3- 2$$
$$a = 1$$
Therefore the solution is: $$a = 1$$ and $$b = 2$$
Solve the equations using graphical method: $$2x- y = 2$$ and $$-4x- y = -4$$
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(1, -1)
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(1, 0)
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(2, -1)
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(0, 1)
Explanation
$$2x- y = 2$$ ----- (1)
$$-4x- y = -4$$ ----- (2)
From equation (1)
$$y = 2x - 2$$ ------ (3)
Assume the value of x = 1, 2 and put those values in equation (3)
If $$x = 1, y = 2(1)- 2 = 0$$
If $$x = 2, y = 2(2) - 2=2$$
Now plotting (1, 0), (2, 2) and joining them, we get a straight line.
From equation (2),
$$-4x- y = -4$$
$$y = -4x + 4$$ ------ (4)
Assume the value of $$x = 1, 2$$ and put those values in equation (3)
If $$x = 1, y =-4(1)+4= 0$$
If $$x = 2, y = -4(1)+4 = -2$$
Plotting (1, 0), (2, -2) and joining them, we get another straight line.
These lines intersect at the point (1, 0) and therefore the solution of the equation is $$x = 1, y = 0$$
0:0:1
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