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Algebraic Expressions Test - 14

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Algebraic Expressions Test - 14
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  • Question 1
    1 / -0
    From 8y+2y2\displaystyle 8-y+{ 2y }^{ 2 } take away (y272y) \displaystyle \left( { y }^{ 2 }-7-2y \right) .
    Solution
    Here we have to just subtract y272yy^2-7-2y from 8y+2y28-y+2y^2

    (8y+2y2)(y272y)=8y+2y2y2+7+2y\therefore \displaystyle \left( 8-y+{ 2y }^{ 2 } \right) -\left( { y }^{ 2 }-7-2y \right) =8-y+{ 2y }^{ 2 }-{ y }^{ 2 }+7+2y 

    =(8+7)+(y+2y)+(2y2y2) =\displaystyle \left( 8+7 \right) +\left( -y+2y \right) +\left( { 2y }^{ 2 }-{ y }^{ 2 } \right) 

    =15+y+y2\displaystyle =15+y+{ y }^{ 2 }

    Hence, required value is y2+y+15y^2+y+15.
  • Question 2
    1 / -0
    What should be subtracted from 2a+6b52a+6b-5 to get 3a+2b+3-3a+2b+3?
    Solution
    Let X=3a+2b+3X=-3a+2b+3 and Y=2a+6b5Y=2a+6b-5
    Let ZZ be the required expression
    Now, X=YZX=Y-Z
    =>Z=YX=>Z=Y-X
    Thus,
        2a+6b5(3a+2b+3)2a+6b-5-(-3a+2b+3)
    =2a+6b5+3a2b3=2a+6b-5+3a-2b-3
    =5a+4b8=5a+4b-8
  • Question 3
    1 / -0
    Find the sum of x22y2,2x24xy+5y2,6y2+11xy6x2\displaystyle { x }^{ 2 }-{ 2y }^{ 2 },{ 2x }^{ 2 }-4xy+{ 5y }^{ 2 },{ 6y }^{ 2 }+11xy-{ 6x }^{ 2 }
    Solution
    Required sum =(x22y2)+(2x24xy+5y2)+(6y2+11xy6x2)=\displaystyle { (x }^{ 2 }-{ 2y }^{ 2 })+{ (2x }^{ 2 }-4xy+{ 5y }^{ 2 })+{ (6y }^{ 2 }+11xy-{ 6x }^{ 2 })

    =(x2+2x26x2)+(2y2+5y2+6y2)+(4xy+11xy) =\displaystyle \left( { x }^{ 2 }+{ 2x }^{ 2 }-{ 6x }^{ 2 } \right) +\left( { -2y }^{ 2 }+{ 5y }^{ 2 }+{ 6y }^{ 2 } \right) +\left( -4xy+11xy \right)  (Combining like terms)

    =3x2+9y2+7xy\displaystyle= -{ 3 }x^{ 2 }+{ 9y }^{ 2 }+7xy.
  • Question 4
    1 / -0
    What should be added to 5x2+2xy+y25x^2+2xy+y^2 to get 3x2+4xy3x^2+4xy?
    Solution
    Let A=3x2+4xyA=3x^2+4xy and B=5x2+2xy+y2B=5x^2+2xy+y^2
    Let CC be the required expression
    Now, A=B+CA=B+C
    =>C=AB=>C=A-B
    Thus, 3x2+4xy(5x2+2xy+y2)3x^2+4xy-(5x^2+2xy+y^2)
    =3x2+4xy5x22xyy2=3x^2+4xy-5x^2-2xy-y^2
    =2x2+2xyy2=-2x^2+2xy-y^2
  • Question 5
    1 / -0
    Simplify
    (a32a2+4a5)(a38a+2a2+5)\displaystyle \left ( a^{3}-2a^{2}+4a-5 \right )-\left ( -a^{3}-8a+2a^{2}+5 \right )
    Solution
    332a2+4a5\displaystyle 3^{3}-2a^{2}+4a-5
    ()a3+2a28a+5\displaystyle (-)-a^{3}+2a^{2}-8a+5
    +         -          +           -
    2a34a2+12a10\displaystyle \underline{\overline{2a^{3}-4a^{2}+12a-10}}
  • Question 6
    1 / -0
    Subtract (5x25x7)(5x^{2}-5x-7) from the sum of (x23),(54x)(x^{2}-3),\:(5-4x) and (9+4x2)(9+4x^{2})
    Solution
    (x23)+(54x)+(9+4x2)=5x24x+11(x^2-3)+(5-4x)+(9+4x^2)=5x^2-4x+11
    Now, (5x24x+11)(5x25x7)=x+18(5x^2-4x+11)-(5x^2-5x-7)=x+18
    Option A is correct.

  • Question 7
    1 / -0
    Simplify : 5x[4x{(2x5)3(3x4) } ] \displaystyle 5x-\left[ 4x-\left\{ \left( 2x-5 \right) -3\left( 3x-4 \right)  \right\}  \right] 
    Solution
    5x[4x{(2x5)3(3x4) } ] \displaystyle 5x-\left[ 4x-\left\{ \left( 2x-5 \right) -3\left( 3x-4 \right)  \right\}  \right] 
    =5x4x+[(2x5)3(3x4)]=5x-4x+[(2x-5)-3(3x-4)]
    =5x4x+2x59x+12=5x-4x+2x-5-9x+12
    =76x=7-6x
  • Question 8
    1 / -0
     a[a+{a+b2a(a2b)} b]\displaystyle -a-[a+\{ { a+b-2a-(a-2b)\}  }-b] is equal to
    Solution
    Given,  a[a+{a+b2a(a2b)} b]\displaystyle -a-[a+\{ { a+b-2a-(a-2b)\}  }-b]
    =a[a+{a+b2aa+2b} b]\displaystyle -a-[a+\{ { a+b-2a-a+2b\}  }-b]
    =a[a+{2a+3b} b]\displaystyle =-a-[a+\{ { -2a+3b\}  }-b]
    =a[a2a+3bb] \displaystyle =-a-\left[ a-2a+3b-b \right] 
    =a[a+2b] \displaystyle =-a-\left[ -a+2b \right] 
    =a+a2b\displaystyle =-a+a-2b
    =2b\displaystyle =-2b, which is simplified form of the given expression.
  • Question 9
    1 / -0
    By how much is x44x2y2+y4\displaystyle x^{4}-4x^{2}y^{2}+y^{4} less than x4+8x2y2+y4\displaystyle x^{4}+8x^{2}y^{2}+y^{4}?
    Solution
    We have to find solution of
    (x4+8x2y2+y4)( x^{4}+8x^{2}y^{2}+y^{4}) (x44x2y2+y4)-(x^{4}-4x^{2}y^{2}+y^{4})
    Separating like terms and unlike terms, we get
    == x4x4+y4y4+8x2y2(4x2y2)x^4 - x^4 +y^4 - y^4 +8x^2y^2-(-4x^2y^2)
    == 8x2y2+4x2y28x^2y^2 + 4x^2y^2
    == 12x2y212x^2y^2
  • Question 10
    1 / -0
    Subtract the sum of (5x27x+4) \displaystyle \left( { 5x }^{ 2 }-7x+4 \right)  and (2x5x3+1) \displaystyle \left( 2x-{ 5x }^{ 3 }+1 \right)  from (3x21+5x) \displaystyle \left( { 3x }^{ 2 }-1+5x \right) .
    Solution
    3x21+5x[(5x27x+4)+(2x5x3+1)]3x^2 - 1 + 5x - [(5x^2-7x+4) + (2x-5x^3+1)]
    = 3x21+5x[5x25x+55x3]3x^2 - 1 +5x - [5x^2-5x+5- 5x^3]
    = 3x21+5x5x2+5x5+5x33x^2 - 1 +5x - 5x^2+5x-5 +5x^3
    =2x2+10x6+5x3 -2x^2 + 10x -6 + 5x^3
    = 5x32x2+10x65x^3 -2x^2 + 10x - 6
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