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CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 3 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Linear Equations
Quiz 3
Solve the following simultaneous equations:
$$37x+29y=13;\, \, 29x+37y=53$$
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$$x=1$$, $$y=4$$
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$$x=3$$, $$y=-1$$
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$$x=-2$$, $$y=3$$
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$$x=1$$, $$y=-2$$
Explanation
Consider the equations:
$$37x+29y=13........(1)$$ and
$$29x+37y=53........(2)$$
Multiplying equation 1 by $$29$$ and
equation 2 by $$37$$, we get:
$$1073x+841y=377........(3)$$ and
$$1073x+1369y=1961........(4)$$
Now subtracting equations 1 from equation 2 we get:
$$1073x+1369y-1073x-841y=1961-377$$
or
$$528y=1584$$
or
$$y=3$$
Substitute the value of $$y$$ in equation 1:
$$37x+87=13$$
$$37x=-74$$
$$x=-2$$
Hence, $$x=-2,y=3$$.
Find the values of the $$x+y$$ and $$x-y$$ from the examples given below without solving for x and y
$$4x+3y=24;\, \, 3x+4y=25$$
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$$7$$ , $$-1$$
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$$1$$ , $$-7$$
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$$1$$ , $$7$$
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$$-7$$ , $$-1$$
Explanation
$$\text{The given equations are}$$
$$4x+3y=24$$ ...(i)
$$3x+4y=25$$ ...(ii)
On adding (i) and (ii), we get
$$7x+7y=49\Rightarrow x+y=7$$
On subtracting (ii) from (i), we get
$$x-y=-1$$
$$\therefore x+y=7 \space and \space x-y= -1$$
Find the values of the $$x+y$$ and $$x-y$$ from the examples given below without solving for x and y
$$7x-5y=-1; \, \, 5x-7y=-11$$
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$$- 1$$, $$-5$$
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$$1$$, $$-5$$
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$$5$$, $$1$$
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$$ 5 $$, $$ -1 $$
Explanation
$$\text{The given equations are}$$
$$7x-5y=-1$$ ...(i)
$$5x-7y=-11$$ ...(ii)
On adding (i) and (ii), we get
$$12x-12y=-12\Rightarrow x-y=-1$$
On subtracting (ii) from (i), we get
$$2x+2y=10\Rightarrow x+y=5$$
$$\therefore x+y=5 \space and \space x-y=-1$$
Solve the following simultaneous equations.
$$3x+4y=18; \, \, 4x+3y=17$$
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$$x = 2$$, $$y = 3$$
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$$x = -2$$, $$y = 5$$
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$$x=4$$, $$y=3$$
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$$x=-2$$, $$y=-3$$
Explanation
The given equations are
$$3x+4y=18$$ ...(i)
$$4x+3y=17$$ ...(ii)
On multiplying (i) by 4 and (ii) by 3 and subtracting, we get
$$12x+16y=72$$
$$\underline {\underset {-}12x\underset {-}{+}9y=\underset{-}51}$$
$$7y=21$$
$$\Rightarrow y=3$$
On putting $$y=3$$ in (i), we get
$$3x+12=18\Rightarrow 3x=6\Rightarrow x=2$$
$$\therefore x=2,y=3$$
Solve the following simultaneous equations:
$$15x-17y=28;\, \, 15y-17x+36=0$$
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$$x=3$$, $$y=1$$
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$$x=-3$$, $$y=1$$
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$$x=6$$, $$y=2$$
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$$x=2$$, $$y=3$$
Explanation
The given equations are
$$15x-17y=28$$ ...(i)
$$17x-15y=36$$ ...(ii)
On multiplying (i) by 17 and (ii) by 15 and subtracting, we get
$$255x-289y=476$$
$$\underline {\underset {-}255x\underset {+}{-}225y=\underset{-}540}$$
$$-64y=-64$$
$$\Rightarrow y=1$$
On putting $$y=1$$ in (i), we get
$$15x-17=28\Rightarrow 15x=45\Rightarrow x=3$$
$$\therefore x=3,y=1$$
Solve following pair of equations by equating the coefficient method:
$$2x\, -\, y\, =\, 9$$
$$3x\, -\, 7y\, =\, 19$$
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$$x=2$$ , $$y=1$$
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$$x=4$$ , $$y=-1$$
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$$x=3$$ , $$y=2$$
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$$x=7$$ , $$y=-8$$
Explanation
Multiply the equation $$2x-y=9$$ by $$7$$ to make the coefficients of $$y$$ equal. Then we get the equations:
$$14x-7y=63.........(1)$$
$$3x-7y=19.........(2)$$
Subtract Equation (2) from Equation (1) to eliminate $$y$$, because the coefficients of $$y$$ are the same. So, we get
$$(14x-3x)+(-7y+7y)=63-19$$
i.e. $$11x = 44$$
i.e. $$x = 4$$
Substituting this value of $$x$$ in (2), we get
$$12-7y=19$$
i.e.
$$-7y=19-12$$
i.e.
$$-7y=7$$
i.e.
$$y=-1$$
Hence, the solution of the equations is $$x = 4, y = -1$$.
Solve the following simultaneous equation:
$$4x+5y = 10; \, \, 5x+4y=17$$
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$$x=3\, \, y=2$$
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$$x=1\, \, y=4$$
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$$x=0\, \, y=3$$
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$$x=5\, \, y=-2$$
Explanation
$$4x+5y=10$$ .....(1)
$$5x+4y=17$$ .......(2)
Adding (1) and (2). we get
$$9x+9y=27$$
$$\Rightarrow x+y=3$$ ......(3)
Now, subtracting (1) from (2). we get,
$$x-y=7$$ .......(4)
Adding (3) and (4). we get
$$2x=10$$
$$\Rightarrow x=5$$
Now, subtracting (3) from (4). we get,
$$-2y=4$$
$$\Rightarrow y=-2$$
$$\therefore x=5 , y=-2$$
Option D is correct.
Solve the given pair of equations by substitution method:
$$x\, +\, 5y\, =\, 18$$
$$3x\, +\, 2y\, =\, 41$$
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$$(13 , 1)$$
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$$(1 , 3)$$
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$$(12 , 17)$$
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$$(9 , 16)$$
Explanation
We have, $$x+5y =18\Rightarrow x = 18-5y$$ ..........(1)
and $$3x+2y=41$$.......(2)
Now substitute value x in terms of y from (1) to (2)
$$\Rightarrow 3(18-5y)+2y=41$$
$$\Rightarrow 54-15y+2y=41$$
$$\Rightarrow 15y-2y=54-41$$
$$\Rightarrow 13y=13\Rightarrow y=\dfrac{13}{13}=1$$
So $$x =18-5y=18-5(1)=13$$
Hence solution set is $$(x,y)=(13,1)$$
Solve the given pair of equations by substitution method:
$$x\, +\, y\, =\, 11$$
$$x\, -\, y\, =\, -3$$
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$$(4 , 6)$$
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$$(3 , 11)$$
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$$(4 , 7)$$
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$$(6 , 2)$$
Explanation
We pick either of the equations and write one variable in terms of the other.
Let us consider the equation
$$x-y = -3$$
and write it as
$$x = -3+y...........(1)$$
Substitute the value of $$x$$ in Equation
$$x+y = 11$$
. We get
$$-3+y+y = 11$$
i.e.
$$-3+y+y = 11$$
i.e.
$$-3+2y = 11$$
i.e.
$$2y = 14$$
Therefore, $$y = 7$$
Substituting this value of $$y$$ in Equation (1), we get
$$x = -3+7=4$$
Hence, the solution is $$x = 4, y = 7$$.
Solve the given pair of equations by substitution method:
$$4a\, -\, b\, =\, 10$$
$$2a\, +\, 3b\, =\, 12$$
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$$a = 0, b = 5$$
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$$a = 7, b = 6$$
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$$a = 1, b = 7$$
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$$a = 3, b = 2$$
Explanation
We have, $$4a-b=10\Rightarrow b=4a-10$$.........(1)
and $$2a+3b=12$$........(2)
Now substitute value of b in terms of a in (2)
$$\Rightarrow 2a+3(4a-10)=12$$
$$\Rightarrow 2a+12a-30=12$$
$$\Rightarrow 14a=12+30=42\Rightarrow a=\dfrac{42}{14}=3$$
So $$b=4a-10=4(3)-10=12-10=2$$
Hence option 'D' is correct choice
Solve following pair of equations by equating the coefficient method:
$$8x\, =\, 5y$$
$$13x\, =\, 8y\, +\, 1$$
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$$x=3$$ and $$y=1$$
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$$x=5$$ and $$y=8$$
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$$x=6$$ and $$y=2$$
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$$x=9$$ and $$y=-5$$
Explanation
Rewriting the given equations, we get
$$8x-5y=0$$ ...(i)
$$13x-8y=1$$ ...(ii)
On multiplying (i) by 13 and (ii) by 8, we get
$$104x-65y=0$$
$$\underline {\underset {-}104x\underset {+}{-}64y=\underset{-}8}$$
$$-y=-8$$
$$\therefore y=8$$
On putting $$y=8$$ in (i), we get
$$x=\frac{5y}8\Rightarrow x=5$$
$$\therefore x=5,y=8$$
Solve the given pair of equations by substitution method:
$$x\, -\, 4y\, =\, -8$$
$$x\, -\, 2y\, =\, 0$$
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$$(2 , -1)$$
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$$(7, -6)$$
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$$(8 , 4)$$
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$$(-3 , 6)$$
Explanation
We have, $$x-4y =-8\Rightarrow x = 4y-8$$ ..........(1)
and $$x-2y=0$$.......(2)
Now substitute value of x in terms of y from (1) to (2)
$$\Rightarrow 4y-8-2y=0$$
$$\Rightarrow 2y=8$$
$$\Rightarrow y=\dfrac{8}{2}=4$$
So $$x =4y-8=4(4)-8=8$$
Hence solution set is $$(x,y)=(8,4)$$
Solve the given pair of equations by substitution method:
$$2a\, +\, 3b\, =\, 6$$
$$3a\, +\, 5b\, =\, 15$$
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$$a = 3, b = 2$$
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$$a = -15, b = 12$$
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$$a = 8, b = 15$$
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$$a = -7, b = 3$$
Explanation
We pick either of the equations and write one variable in terms of the other.
Let us consider the equation
$$2a+3b=6$$
and write it as
$$b=\frac { 6-2a }{ 3 } .........(1)$$
Substitute the value of $$b$$ in Equation
$$3a+5b=15$$
. We get
$$3a+5\left( \frac { 6-2a }{ 3 } \right) =15\\ \Rightarrow 9a+5\left( 6-2a \right) =45\\ \Rightarrow 9a+30-10a=45\\ \Rightarrow -a=45-30\\ \Rightarrow a=-15$$
Therefore, $$a=-15$$
Substituting this value of $$a$$ in Equation (1), we get
$$b=\frac { 6-(-30) }{ 3 }$$
$$b=\frac { 36 }{ 3 }=12$$
Hence, the solution is $$a=-15, b=12$$.
Solve the following pairs of equations:
$$ \displaystyle \dfrac{x+1}{4}= \displaystyle \dfrac{2}{3}\left ( 1-2y \right ) $$
$$ \displaystyle \dfrac{2+5y}{3}= \displaystyle \dfrac{x}{7} -2 $$
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$$ x= 8,\:y= -1 $$
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$$ x= 3,\:y= -1 $$
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$$ x= 7,\:y= -1 $$
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$$ x= 9,\:y= -1 $$
Solve the given pair of equations by substitution method:
$$x\, +\, y\, =\, 0$$
$$y\, -\, x\, =\, 6$$
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$$(2 , 7)$$
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$$(-3 , 3)$$
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$$(4 , 9)$$
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$$(-6 , 2)$$
Explanation
We have, $$x+y =0\Rightarrow x = -y$$ ..........(1)
and $$y-x=6$$.......(2)
Now substitute value of x in terms of y from (1) to (2)
$$\Rightarrow y-(-y)=6$$
$$\Rightarrow y+y=6$$
$$\Rightarrow 2y=6$$
$$\Rightarrow y=\dfrac{6}{2}=3$$
$$x =-y=-(3)=-3$$
Hence solution set is $$(x,y)=(-3,3)$$
Solve the following pair of simultaneous equations:
$$\displaystyle \frac{x}{3}\, +\, \frac{x\, +\, y}{6}\, =\,3;\, \frac{y}{3}\, -\, \frac{x\, -\, y}{2}\, =\, 6$$
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$$x=2,y=-5$$
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$$x=7,y=4$$
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$$x=6,y=3$$
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$$x=3,y=9$$
Explanation
$$\frac{x}{3}+\frac{x+y}{6}=3\Rightarrow \frac{2x+x+y}6=3\Rightarrow 3x+y=18 $$ ...(i)
$$\frac{y}{3}-\frac{x-y}{2}=6\Rightarrow \frac{2y-3x+3y}6=6\Rightarrow -3x+5y=36$$ ...(ii)
On adding (i) and (ii), we get
$$3x+y=18$$
$$\underline {-3x+5y=36}$$
$$6y=54$$
$$\therefore y=9$$
On putting $$y=9$$ in (i), we get
$$3x+9=18\Rightarrow 3x=9\Rightarrow x=3$$
$$\therefore x=3,y=9$$
Solve the following pair of simultaneous equations:
$$4x\, -\, 3y\, =\, 8$$
$$3x\, -\, 4y\, =\, - 1$$
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$$x=3,y=-1$$
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$$x=5,y=4$$
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$$x=-4,y=2$$
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$$x=7,y=-6$$
Explanation
Multiply the equation $$4x-3y=8$$ by $$3$$ and
equation $$3x-4y=-1$$ by $$4$$
to make the coefficients of $$x$$ equal. Then we get the equations:
$$12x-9y=24.........(1)$$
$$12x-16y=-4.........(2)$$
Subtract Equation (2) from Equation (1) to eliminate $$x$$, because the coefficients of $$x$$ are the same. So, we get
$$(12x-12x)+(-9y+16y)=24+4$$
i.e. $$7y =28$$
i.e. $$y =4$$
Substituting this value of $$y$$ in
the equation $$4x-3y=8$$
, we get
$$4x-12=8$$
i.e.
$$4x=20$$
i.e.
$$x=5$$
Hence, the solution of the equations is $$x = 5, y =4$$.
Solve the following pair of simultaneous equations:
$$8a\, -\, 7b\, =\, 1$$
$$4a\, =\, 3b\, +\, 5$$
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$$a=8 , b=9$$
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$$a=3 , b=-8$$
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$$a=7 , b=-6$$
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$$a=1 , b=-1$$
Explanation
Multiply the equation $$4a=3b+5$$ or
$$4a-3b=5$$
by $$2$$
to make the coefficients of $$a$$ equal. Then we get the equations:
$$8a-7b=1.........(1)$$
$$8a-6b=10.........(2)$$
Subtract Equation (2) from Equation (1) to eliminate $$a$$, because the coefficients of $$a$$ are the same. So, we get
$$(8a-8a)+(-7b+6b)=1-10$$
i.e. $$-b=-9$$
i.e. $$b=9$$
Substituting this value of $$b$$ in
the equation $$8a-7b=1$$
, we get
$$8a-63=1$$
i.e.
$$8a=64$$
i.e.
$$a=8$$
Hence, the solution of the equations is $$a=8, b=9$$.
Solve the following pair of linear (simultaneous) equations by the method of elimination:
$$y= 4x-7$$
$$16x-5y= 25$$
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$$\displaystyle x= \displaystyle \frac{3}{2},\displaystyle y= 6$$
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$$\displaystyle x= \displaystyle \frac{5}{2},\displaystyle y= 3$$
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$$\displaystyle x= \displaystyle \frac{7}{4},\displaystyle y= 2$$
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$$\displaystyle x= \displaystyle \frac{9}{2},\displaystyle y= 0$$
Explanation
$$4x - y = 7$$ ...$$(1)$$
$$16x - 5y = 25$$ ...$$(2)$$
Multiplying $$(1)$$ by $$4$$, we get
$$16x - 4y = 28$$ ...$$(3)$$
$$16x - 5y = 25$$ ...$$(2)$$
Subtracting $$(3)$$ and $$(2)$$,
$$y = 3$$
Substituting $$y = 3$$ in $$(1)$$, we get $$x$$ as
$$4x - 3 = 7$$
$$4x = 10$$
$$x = \dfrac{10}{4}$$ or $$x = \dfrac{5}{2}$$
$$x = \dfrac{5}{2}, y = 3$$
Option $$B$$
Solve the following pair of simultaneous equations:
$$5x\, -\, 6y\, =\, 8$$
$$7y\, -\, 15x\,=\, 9$$
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(1 , 3)
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(6 , -1)
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(-2 , -3)
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(3 , -5)
Explanation
Multiply the equation $$5x-6y=8$$ by $$3$$
to make the coefficients of $$x$$ equal.
Another
equation is $$7y-15x=9$$ or
$$-15x+7y=9$$.
Then we get the equations:
$$15x-18y=24.........(1)$$
$$-15x+7y=9.........(2)$$
Add Equations 1 and 2 to eliminate $$x$$, because the coefficients of $$x$$ are the same. So, we get
$$(15x-15x)+(-18y+7y)=24+9$$
i.e. $$-11y=33$$
i.e. $$y=-3$$
Substituting this value of $$y$$ in
the equation
$$5x-6y=8$$
, we get
$$5x+18=8$$
i.e.
$$5x=-10$$
i.e.
$$x=-2$$
Hence, the solution of the equations is $$x = -2, y =-3$$.
Solve following pair of equations by equating the coefficient method:
$$3x\, -\, 7y\, =\, 35$$
$$2x\, +\, 5y\, =\, 4$$
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$$x=5 , y=-1$$
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$$x=4 , y=- 1$$
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$$x=5 , y=-7$$
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$$x=7,y=-2$$
Explanation
Multiply the equation $$3x-7y=35$$ by $$2$$ and
equation $$2x+5y=4$$ by $$3$$
to make the coefficients of $$x$$ equal. Then we get the equations:
$$6x-14y=70.........(1)$$
$$6x+15y=12.........(2)$$
Subtract Equation (2) from Equation (1) to eliminate $$x$$, because the coefficients of $$x$$ are the same. So, we get
$$(6x-6x)+(-14y-15y)=70-12$$
i.e. $$-29y =58$$
i.e. $$y = -2$$
Substituting this value of $$y$$ in (2), we get
$$6x-30=12$$
i.e.
$$6x=42$$
i.e.
$$x=7$$
Hence, the solution of the equations is $$x = 7, y = -2$$.
Solve following pair of equations by equating the coefficient method:
$$x\, +\, 2y\, =\, 11$$
$$2x\, -\, y\, =\, 2$$
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$$x=3 , y=4$$
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$$x=5 , y=4$$
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$$x=-1 , y=-1$$
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$$x=6 , y=-4$$
Explanation
Multiply the equation $$x+2y=11$$ by $$2$$ to make the coefficients of $$x$$ equal. Then we get the equations:
$$2x+4y=22.........(1)$$
$$2x-y=2.........(2)$$
Subtract Equation (2) from Equation (1) to eliminate $$x$$, because the coefficients of $$x$$ are the same. So, we get
$$(2x-2x)+(4y+y)=22-2$$
i.e. $$5y =20$$
i.e. $$y = 4$$
Substituting this value of $$y$$ in (2), we get
$$2x-4=2$$
i.e.
$$2x=6$$
i.e.
$$x=3$$
Hence, the solution of the equations is $$x = 3, y = 4$$.
Solve the following pair of simultaneous equations:
$$\displaystyle \frac{x}{3}\, =\, \frac{y}{2}\,;\, \frac{2x}{3}\, -\, \frac{y}{2}\, =\, 2$$
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$$(-3 , 4)$$
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$$(1 , 9)$$
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$$(6 , 4)$$
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$$(3 , 8)$$
Explanation
The equation
$$\frac { x }{ 3 } =\frac { y }{ 2 }$$
can be rewritten as:
$$\frac { x }{ 3 } =\frac { y }{ 2 } \\ \Rightarrow 2x=3y\\ \Rightarrow 2x-3y=0\quad .........(1)$$
T
he equation
$$\frac { 2x }{ 3 } -\frac { y }{ 2 }=2$$
can be solved as:
$$\frac { 2x }{ 3 } -\frac { y }{ 2 } =2\\ \Rightarrow \frac { 4x-3y }{ 6 } =2\\ \Rightarrow 4x-3y=12\quad .........(2)$$
Subtract Equation 2 from equation 1 to eliminate $$y$$, because the coefficients of $$y$$ are the same. So, we get
$$(4x-2x)+(-3y+3y)=12-0$$
i.e. $$2x=12$$
i.e. $$x=6$$
Substituting this value of $$x$$ in
the equation
1
, we get
$$12-3y=0$$
i.e.
$$3y=12$$
i.e.
$$y=4$$
Hence, the solution of the equations is $$x =6, y =4$$.
Solve the following pair of simultaneous equations:
$$\displaystyle\frac{x}{2}\, -\,\frac{y}{3}\, =\, 2\,;\, \frac{x}{5}\, +\, \frac{y}{3}\, =\, 15$$
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$$x = \displaystyle 24\frac{2}{7}, y = \displaystyle 30\frac{3}{7}$$
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$$x = \displaystyle 12\frac{1}{3}, y = \displaystyle 10\frac{2}{3}$$
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$$x = \displaystyle 16\frac{7}{6}, y = \displaystyle 13\frac{4}{3}$$
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$$x = \displaystyle 2\frac{22}{7}, y = \displaystyle 12\frac{13}{17}$$
Explanation
The equation
$$\cfrac { x }{ 2 } -\frac { y }{ 3 } =2$$
can be solved as:
$$\cfrac { x }{ 2 } -\cfrac { y }{ 3 } =2\\ \Rightarrow \cfrac { 3x-2y }{ 6 } =2\\ \Rightarrow 3x-2y=12\quad .........(1)$$
T
he equation
$$\cfrac { x }{ 5 } +\cfrac { y }{ 3 } =15$$
can be solved as:
$$\cfrac { x }{ 5 } +\cfrac { y }{ 3 } =15\\ \Rightarrow \cfrac { 3x+5y }{ 15 } =15\\ \Rightarrow 3x+5y=225\quad .........(2)$$
Subtract Equation 2 from equation 1 to eliminate $$x$$, because the coefficients of $$x$$ are the same. So, we get
$$(3x-3x)+(-2y-5y)=12-225$$
i.e. $$-7y=-213$$
i.e.
$$y=\cfrac { 213 }{ 7 } =30\cfrac { 3 }{ 7 }$$
Substituting this value of $$y$$ in
the equation
1
, we get
$$3x-2\left( \cfrac { 213 }{ 7 } \right) =12\\ \Rightarrow 3x-\cfrac { 426 }{ 7 } =12\\ \Rightarrow 21x-426=84\\ \Rightarrow 21x=510\\ \Rightarrow x=\cfrac { 510 }{ 21 } =\cfrac { 170 }{ 7 } =24\cfrac { 2 }{ 7 }$$
Hence, the solution of the equations is
$$x=24\cfrac { 2 }{ 7 } ,y=30\cfrac { 3 }{ 7 }$$
.
Solve for $$x$$ and $$y$$:
$$4x= 17-\displaystyle \frac{x-y}{8}$$
$$2y+x= 2+\displaystyle \frac{5y+2}{3}$$
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$$x= 4;y= -4$$
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$$x= 1;y= -6$$
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$$x= -6;y= -3$$
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$$x= 2;y= -7$$
Explanation
Given equations are,
$$4x=17-\dfrac {x-y}{8}$$
$$\Rightarrow 32x=136-x+y$$
$$\Rightarrow 33x-y=136$$ ....(1)
and $$2y+x=2+\dfrac {5y+2}{3}$$
$$\Rightarrow 3(2y+x)=6+5y+2$$
$$\Rightarrow 6y+3x=6+5y+2$$
$$\Rightarrow 3x+y=8$$ ....(2)
Add equations (1) and (2), we get
$$36x=144$$
$$\Rightarrow x=4$$
Put this value in equation (1), we get
$$33(4)-y=136$$
$$\Rightarrow 132-y=136$$
$$\Rightarrow y=-4$$
Therefore, the solution is $$x=4, y=-4$$.
Solve the following pair of simultaneous equations:
$$3x\, +\, 5(y\, +\, 2)\, =\, 1$$
$$3x\, +\, 8y\, =\, 0$$
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$$(-8 , 3)$$
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$$(-4 , 9)$$
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$$(1 , -3)$$
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$$(0 , -9)$$
Explanation
The equation $$3x+5(y+2)=1$$ can be written as
$$3x+5y+10=1$$ or
$$3x+5y=-9$$
.
Another
equation is $$3x+8y=0$$.
Then we get the equations:
$$3x+5y=-9.........(1)$$
$$3x+8y=0.........(2)$$
Subtract Equation 2 from equation 1 to eliminate $$x$$, because the coefficients of $$x$$ are the same. So, we get
$$(3x-3x)+(5y-8y)=-9-0$$
i.e. $$-3y=-9$$
i.e. $$y=3$$
Substituting this value of $$y$$ in
the equation
$$3x+8y=0$$
, we get
$$3x+24=0$$
i.e.
$$3x=-24$$
i.e.
$$x=-8$$
Hence, the solution of the equations is $$x = -8, y =3$$.
Solve the following pair of simultaneous equations
$$3x\, +\,2y\, =\, -1$$
$$6y\, =\, 5(1\, -\, x)$$
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$$x=- 2$$ and $$y=2.5$$
0%
$$x=- 1$$ and $$y=3$$
0%
$$x=6$$ and $$y=1.5$$
0%
$$x=- 7$$ and $$y=-4$$
Explanation
$$3x+2y=-1$$ ...(i)
$$6y=5(1 - x)\Rightarrow 5x+6y=5$$ ...(ii)
On multiplying (i) by 5 and (ii) by 3, we get
$$15x+10y=-5$$
$$\underline {\underset {-}15x\underset {-}{+}18y=\underset{-}15}$$
$$-8y=-20$$
$$\therefore y=\frac{20}8=2.5$$
On putting $$y=2.5$$ in (i), we get
$$3x+2(2.5)=-1\Rightarrow 3x=-6\Rightarrow x=-2$$
$$\therefore x=-2,y=2.5$$
Solve the following pair of simultaneous equations:
$$\displaystyle \frac{x\, -\, 1}{2}\, +\, \frac{y\, +\, 1}{5}\, =\, 4\frac{1}{5}\,;\, \frac{x\, +\,y}{3}\, =\, y\, -\, 1$$
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0%
$$x=7,y=5$$
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$$x=2,y=4$$
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$$x=0,y=6$$
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$$x=5,y=3$$
Explanation
$$\displaystyle \frac{x-1}{2}+ \frac{y+1}{5}=4\frac{1}{5}\Rightarrow \frac{5x-5+2y+2}{10}=\frac{21}5\Rightarrow 5x+2y=45$$ ...(i)
$$\displaystyle \frac{x+y}{3}=y-1\Rightarrow x+y=3y-3\Rightarrow x-2y=-3$$ ...(ii)
On multiplying (i) by 1 and (ii) by 5, we get
$$5x+2y=45$$
$$\underline {\underset {-}5x\underset {+}{-}10y=\underset{+}{-}15}$$
$$12y=60$$
$$\therefore y=5$$
On putting $$y=5$$ in (ii), we get
$$x-2(5)=-3\Rightarrow x=7$$
$$\therefore x=7,y=5$$
Find two numbers such that twice of the first added to the second gives $$21$$, and twice the second added to the first gives $$27$$.
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$$5$$ and $$11$$
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$$9$$ and $$16$$
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$$3$$ and $$18$$
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$$3$$ and $$17$$
Explanation
Let the first number is $$x$$ and second is $$y$$
Given twice of first added with second gives $$21$$
$$\therefore 2x+y=21$$ ...$$(1)$$
And given twice of second added with first gives $$27$$
$$\therefore x+2y=27$$ ...$$(2)$$
Multiply $$(1)$$ by $$2$$
Then $$4x+2y=42$$ ...$$(3)$$
Subtract $$(3)$$ with $$(2)$$
Then $$3x=15$$
Or $$x=5$$
Put the value of $$x=5$$ in $$(1)$$
Then $$10+y=21$$
Or $$y=21-10$$
Or $$y=11$$
Then first number is $$5$$ and second is $$11$$.
A man's age is three times that of his son and in twelve years he will be twice as old as his son would be. What are their present ages?
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Present age of man is $$60$$ years and
Present age of son is $$20$$ years
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Present age of man is $$45$$ years and
Present age of son is $$15$$ years
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Present age of man is $$54$$ years and
Present age of son is $$18$$ years
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Present age of man is $$36$$ years and
Present age of son is $$12$$ years
Explanation
Let the present age of son be $$x$$ years
As per question the
present
age of father is three times
then
present age of father is $$3x$$
As per question after $$12$$ years the age of father will be twice the age of son
$$3x+12=2(x+12)$$
$$\Rightarrow 3x+12=2x+24$$
$$\Rightarrow 3x-2x=24-12$$
$$\Rightarrow x=12$$
Then present age of son is $$12$$ years
So
present age of father $$=3\times12=36$$
years
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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