If the system of equations is inconsistent, then k =
Question: If the system of equations is inconsistent, thenk= $3 x+y=1$ $(2 k-1) x+(k-1) y=2 k+1$ (a) 1 (b) 0 (c) $-1$ (d) 2 Solution: The given system of equations is inconsistent, $3 x+y=1$ $(2 k-1) x+(k-1) y=2 k+1$ If the system of equations is in consistent, we have $a_{1} b_{2}-a_{2} b_{1}=0$ $3 \times(k-1)-1(2 k-1)=0$ $3 k-3-2 k+1=0$ $1 k-2=0$ $1 k=2$ Therefore, the value of $k$ is 2 . Hence, the correct choice is $d$....
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Question: Find the value $x^{3}-2 x^{2} b+3 x y^{2}-6 y^{3}$ Solution: Taking $x^{2}$ common in $\left(x^{3}-2 x^{2} y\right)$ and $+3 y^{2}$ common in $\left(3 x y^{2}-6 y^{3}\right)$ $=x^{2}(x-2 y)+3 y^{2}(x-2 y)$ Taking (x - 2y) common in the terms $=(x-2 y)\left(x^{2}+3 y^{2}\right)$ $\therefore x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3}=(x-2 y)\left(x^{2}+3 y^{2}\right)$...
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Question: Find the value $x^{2}+y-x y-x$ Solution: On rearranging $x^{2}-x y-x+y$ Taking $x$ common in the $\left(x^{2}-x y\right)$ and $-1$ in $(-x+y)$ $=x(x-y)-1(x-y)$ Taking $(x-y)$ common in the terms $=(x-y)(x-1)$ $\therefore x^{2}+y-x y-x=(x-y)(x-1)$...
Read More →If the system of equations has infinitely many solutions, then
Question: If the system of equations has infinitely many solutions, then $2 x+3 y=7$ $(a+b) x+(2 a-b) y=21$ (a) $a=1, b=5$ (b) $a=5, b=1$ (c) $a=-1, b=5$ (d) $a=5, b=-1$ Solution: The given systems of equations are $2 x+3 y=7$ $(a+b) x+(2 a-b) y=21$ For the equations to have infinite number of solutions, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Here $a_{1}=2, a_{2}=(a+b), b_{1}=3, b_{2}=2 a-b, c_{1}=7, c_{2}=21$ $\frac{2}{a+b}=\frac{3}{2 a-b}=\frac{7}{21}$ Let us take $\frac...
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Question: Find the value $a^{2} x^{2}+\left(a x^{2}+1\right) x+a$ Solution: We multiply $x\left(a x^{2}+1\right)=a x^{3}+x$ $=a^{2} x^{2}+a x^{3}+x+a$ Taking common $a x^{2}$ in $\left(a^{2} x^{2}+a x^{3}\right)$ and 1 in $(x+a)$ $=a x^{2}(a+x)+1(x+a)$ $=a x^{2}(a+x)+1(a+x)$ Taking (a + x) common in both the terms $=(a+x)\left(a x^{2}+1\right)$ $\therefore a^{2} x^{2}+\left(a x^{2}+1\right) x+a=(a+x)\left(a x^{2}+1\right)$...
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Question: Find the value $x\left(x^{3}-y^{3}\right)+3 x y(x-y)$ Solution: Elaborating $x^{3}-y^{3}$ using the identity $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$ $=x(x-y)\left(x^{2}+x y+y^{2}\right)+3 x y(x-y)$ Taking common x(x - y) in both the terms $=x(x-y)\left(x^{2}+x y+y^{2}+3 y\right)$ $\therefore x\left(x^{3}-y^{3}\right)+3 x y(x-y)$ $=x(x-y)\left(x^{2}+x y+y^{2}+3 y\right)$...
Read More →The value of k for which the system of equations 3x + 5y = 0
Question: The value ofkfor which the system of equations 3x+ 5y= 0 andkx+ 10y= 0 has non-zero solution, is (a) 0 (b) 2 (c) 6 (d) 8 Solution: The given system of equations are, $3 x+5 y=0$ $k x+10 y=0$ Here, $a_{1}=3, a_{2}=k, b_{1}=5, b_{2}=10$ $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b} \neq 0$ $\frac{3}{k}=\frac{5}{10} \neq 0$ By cross multiply we get $30=5 k$ $\frac{30}{5}=k$ $6=k$ Therefore the value of k is 6, Hence, the correct choice is c....
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Question: Find the value $a(a+b)^{3}-3 a^{2} b(a+b)$ Solution: Taking (a + b) common in the two terms $=(a+b)\left\{a(a+b)^{2}-3 a^{2} b\right\}$ Now, using $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $=(a+b)\left\{a\left(a^{2}+b^{2}+2 a b\right)-3 a^{2} b\right\}$ $=(a+b)\left\{a^{3}+a b^{2}+2 a^{2} b-3 a^{2} b\right\}$ $=(a+b)\left\{a^{3}+a b^{2}-a^{2} b\right\}$ $=(a+b) p\left\{a^{2}+b^{2}-a b\right\}$ $=p(a+b)\left(a^{2}+b^{2}-a b\right)$ $\therefore a(a+b)^{3}-3 a^{2} b(a+b)$ $=a(a+b)\left(a^{2}+b^{2}-a ...
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Question: Find the value $x^{3}+x-3 x^{2}-3$ Solution: Taking $x$ common in $x^{3}+x$ $=x\left(x^{2}+1\right)-3 x^{2}-3$ Taking $-3$ common in $-3 x^{2}-3$ $=x\left(x^{2}+1\right)-3\left(x^{2}+1\right)$ Now, we take $\left(x^{2}+1\right)$ common $=\left(x^{2}+1\right)(x-3)$ $\therefore x^{3}+x-3 y^{2}-3=\left(x^{2}+1\right)(x-3)$...
Read More →The value of k for which the system of equations x + 2y − 3 = 0
Question: The value of $k$ for which the system of equations $x+2 y-3=0$ and $5 x+k y+7=0$ has no solution, is (a) 10 (b) 6 (c) 3 (d) 1 Solution: The given system of equations are $x+2 y-3=0$ $5 x+k y+7=0$ For the equations to have no solutions, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ $\frac{1}{5}=\frac{2}{k}=\frac{-3}{7}$ If we take $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$ $\frac{1}{5}=\frac{2}{k}$ $k=10$ Therefore the value ofkis10. Hence, correct choice is $a$....
Read More →The value of k for which the system, of equations has infinite number of solutions, is
Question: The value ofkfor which the system, of equations has infinite number of solutions, is $2 x+3 y=5$ $4 x+k y=10$ (a) 1 (b) 3 (c) 6 (d) 0 Solution: The given system of equations are $2 x+3 y=5$ $4 x+k y=10$ For the equations to have infinite number of solutions, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Here, $a_{1}=2, a_{2}=4, b_{1}=3, b_{2}=k$ Therefore $\frac{2}{4}=\frac{3}{k}=\frac{5}{10}$ By cross multiplication of $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$ we get, ...
Read More →The value of k for which the system of equations has a unique solution, is
Question: The value ofkfor which the system of equations has a unique solution, is $k x-y=2$ $6 x-2 y=3$ (a) $=3$ (b) $\neq 3$ (c) $\neq 0$ (d) $=0$ Solution: The given system of equations are $k x-y=2$ $6 x-2 y=3$ $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ for unique solution Here $a_{1}=k, a_{2}=6, b_{1}=-1, b_{2}=-2$ By cross multiply we get $2 k \neq 6$ $k \neq \frac{6}{2}$ $k \neq 3$ Hence, the correct choice is $b$....
Read More →Write the number of solutions of the following pair of linear equations:
Question: Write the number of solutions of the following pair of linear equations: $x+3 y-4=0$ $2 x+6 y=7$ Solution: The given linear pair of equations are $x+3 y-4=0$ $2 x+6 y=7$ $a_{1}=1, a_{2}=2, b_{1}=3, b_{2}=6, c_{1}=4, c_{2}=7$ $\frac{a_{1}}{a_{2}}=\frac{1}{2}$ $\frac{b_{1}}{b_{2}}=\frac{3}{6}$ $\frac{b_{1}}{b}=\frac{1}{2}$ $\frac{c_{1}}{c_{2}}=\frac{4}{7}$ If $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ then $a_{1} b_{2}=a_{2} b_{1}$ $a_{1} b_{3}-a_{2} b_{1}=0$ $1 \...
Read More →Evaluate:
Question: Evaluate: (a) $25^{3}-75^{3}+50^{3}$ (b) $48^{3}-30^{3}-18^{3}$ (c) $(1 / 2)^{3}+(1 / 3)^{3}-(5 / 6)^{3}$ (d) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$ Solution: Given, (a) $25^{3}-75^{3}+50^{3}$ we know that, $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ here, $a=25, b=-75, c=50$ $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ $a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$ $a^{3}+b^{3}+c^{3}=(25-75+50)\...
Read More →Write the number of solution of the following pair of linear equations:
Question: Write the number of solution of the following pair of linear equations: $x+2 y-8=0$ $2 x+4 y=16$ Solution: The given equations are $x+2 y-8=0$ $2 x+4 y-16=0$ $a_{1}=1, a_{2}=2, b_{1}=2, b_{2}=4, c_{1}=8, c_{2}=16$ $\frac{a_{1}}{a_{2}}=\frac{1}{2} ; \frac{b_{1}}{b_{2}}=\frac{2}{4} ; \frac{c_{1}}{c_{2}}=\frac{8}{16}$ $\frac{a_{1}}{a_{2}}=\frac{1}{2} ; \frac{b_{1}}{b_{2}}=\frac{1}{2} ; \frac{c_{1}}{c_{2}}=\frac{1}{2}$ $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Every sol...
Read More →For what value of k, the following pair of linear equation has infinitely many solutions?
Question: For what value ofk, the following pair of linear equation has infinitely many solutions? $10 x+5 y-(k-5)=0$ $20 x+10 y-k=0$ Solution: The given equations are $10 x+5 y-(k-5)=0$ $20 x+10 y-k=0$ $\frac{a_{1}}{a_{2}}=\frac{10}{20}, \frac{b_{1}}{b_{2}}=\frac{5}{10}, \frac{c_{1}}{c_{2}}=\frac{k-5}{k}$ For the equations to have infinite number of solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Let us take $\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ $\frac{5}{10}=\frac{k...
Read More →If a + b + c = 9 and a2 + b2 + c2 = 35, Find the value of a3 + b3 + c3 - 3abc
Question: If $a+b+c=9$ and $a^{2}+b^{2}+c^{2}=35$, Find the value of $a^{3}+b^{3}+c^{3}-3 a b c$ Solution: Given, $a+b+c=9$ and $a^{2}+b^{2}+c^{2}=35$ We know that, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$ $9^{2}=35+2(a b+b c+c a)$ $81=35+2(a b+b c+c a)$ $81-35=2(a b+b c+c a)$ $46 / 2=a b+b c+c a$ $a b+b c+c a=23$ we know that, $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left[\left(a^{2}+b^{2}+c^{2}\right)-(a b+b c+c a)\r...
Read More →Write the set of values of a and b for which the following system of equations has infinitely many solutions.
Question: Write the set of values ofaandbfor which the following system of equations has infinitely many solutions. $2 x+3 y=7$ $2 a x+(a+b) y=28$ Solution: The given equations are $2 x+3 y-7=0$ $2 a x+(a+b) y-28=0$ For the equations to have infinite number of solutions, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Therefore $\frac{2}{2 a}=\frac{3}{a+b}=\frac{7}{28}$ Let us take $\frac{2}{2 a}=\frac{3}{a+b}$ $2(a+b)=2 a \times 3$ $2 a+2 b=6 a$ $0=6 a-2 a-2 b$ $0=4 a-2 b$ $\frac{...
Read More →If a + b + c = 9 and ab + bc + ca = 26, Find the value of
Question: If $a+b+c=9$ and $a b+b c+c a=26$, Find the value of $a^{3}+b^{3}+c^{3}-3 a b c$ Solution: Given, a + b + c = 9 and ab + bc + ca = 26 We know that, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$ $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(26)$ $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+52$ $9^{2}=a^{2}+b^{2}+c^{2}+52$ $81-52=a^{2}+b^{2}+c^{2}$ $a^{2}+b^{2}+c^{2}=29$ we know that, $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)$ $\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left[\left(a^{2}...
Read More →If x + y + z = 8 and xy + yz + zx = 20, Find the value of
Question: If $x+y+z=8$ and $x y+y z+z x=20$, Find the value of $x^{3}+y^{3}+z^{3}-3 x y z$ Solution: Given, x + y + z = 8 and xy + yz + zx = 20 We know that, $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+z x)$ $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(20)$ $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+40$ $8^{2}=x^{2}+y^{2}+z^{2}+40$ $64-40=x^{2}+y^{2}+z^{2}$ $x^{2}+y^{2}+z^{2}=24$ we know that, $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)$ $\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) x^{3}+y^{3}+z^{3}-3 x y z$ $=(x+y+z)\left[\left(x^...
Read More →Find the following products:
Question: Find the following products: (a) $(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$ (b) $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$ (c) $(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$ (d) $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$ Solution: Given, (a) $(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$ we know that, $x^{3}+y^{3}+z^{3}-3 x y z$ $=(x+y+z)\left(x^{2}+y^{2...
Read More →Write the value of k for which the system of equations x + ky = 0,
Question: Write the value of $k$ for which the system of equations $x+k y=0,2 x-y=0$ has unique solution. Solution: The given equations are $x+k y=0$ $2 x-y=0$ $a_{1}=1, a_{2}=2, b_{1}=k, b_{2}=-1$ $\frac{a_{1}}{a_{2}}=\frac{1}{2}$ $\frac{b_{1}}{b_{2}}=\frac{k}{-1}$ For unique solution $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$ $-1 \times 1 \neq 2 \times k$ $-1 \neq 2 k$ $\frac{-1}{2} \neq k$ For all real values of $k$, except $k=\frac{-1}{2}$ the equations have unique solutions....
Read More →Write the value of k for which the system of equations 3x − 2y = 0
Question: Write the value of $k$ for which the system of equations $3 x-2 y=0$ and $k x+5 y=0$ has infinitely may solutions. Solution: The given equations are $3 x-2 y=0$ $k x+5 y=0$ $\frac{a_{1}}{a_{2}}=\frac{3}{k}, \frac{b_{1}}{b_{2}}=\frac{-2}{5}, \frac{c_{1}}{c_{2}}=\frac{0}{0}$ For the equations to have infinite number of solutions, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Therefore, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$ $\frac{3}{k}=\frac{-2}{5}$ By cross multiplic...
Read More →If x = – 2 and y = 1, by using an identity find the value of the following:
Question: If x = 2 and y = 1, by using an identity find the value of the following: (a) $\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$ (b) $(2 / x-x / 2)\left(4 / x^{2}+x^{2} / 4+1\right)$ (c) $(5 y+15 / y)\left(25 y^{2}-75+225 / y^{2}\right)$ Solution: Given, (a) $\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$ We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\rig...
Read More →ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y − 5)°,
Question: $\mathrm{ABCD}$ is a cyclic quadrilateral such that $\angle \mathrm{A}=(4 y+20)^{\circ}, \angle \mathrm{B}=(3 y-5)^{\circ}, \angle \mathrm{C}=(-4 x)^{\circ}$ and $\angle \mathrm{D}=(7 x+5)^{\circ}$. Find the four angles. Solution: We know that the sum of the opposite angles of cyclic quadrilateral is $180^{\circ}$ in the cyclic quadrilateral $A B C D$ angles $A$ and $C$ and angles $B$ and $D$ pairs of opposite angles Therefore $\angle A+\angle C=180^{\circ}$ and $\angle B+\angle D=180^...
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