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Algebraic Expressions Test - 17

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Algebraic Expressions Test - 17
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  • Question 1
    1 / -0
    If $$P=3x-4y-8z$$, $$Q=-10y+7x+11z$$ and $$R=19z-6y+4x$$, then $$P-Q+R$$ is equal to
    Solution
    $$P=3x-4y-8z$$
    $$Q=-10y+7x+11z$$
    $$R=19z-6y+4x$$
    $$\therefore P-Q+R=(3x-4y-8z)-(-10y+7x+11z)+(19z-6y+4x)$$
    $$=3x-4y-8z+10y-7x-11z+19z-6y+4x$$
    $$=3x-7x+4x-4y+10y-6y-8z-11z+19z=7x-7x-10y+10y-19z+19z=0$$
  • Question 2
    1 / -0
    What is the simplified result of following the steps in order?
    $$(1)$$ add $$5y$$ to $$2x$$
    $$(2)$$ multiply the sum by $$3$$
    $$(3)$$ subtract x$$+$$y from the product
    Solution

  • Question 3
    1 / -0
    The sum of three expressions is $${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$$. If two of them are $$4{ x }^{ 2 }-5{ y }^{ 2 }+3{ z }^{ 2 }$$ and $$-3{ x }^{ 2 }+4{ y }^{ 2 }+2{ z }^{ 2 }$$, the third expression is
    Solution
    Add two expressions 

    $$\Rightarrow \left ( 4x^{2}-5y^{2}+3z^{2} \right )+(-3x^{2}+4y^{2}+2z^{2})$$

    $$=\left ( 4x^{2}-5y^{2}+3z^{2}-3x^{2}+4y^{2}+2z^{2} \right )$$

    $$=4x^{2}-3x^{2}-5y^{2}+4y^{2}+3z^{2}+2z^{2}$$

    $$=x^{2}-y^{2}+5z^{2}$$

    But $$x^{2}-y^{2}+5z^{2}+P(x)=x^{2}+y^{2}+z^{2}$$, where $$P(x)$$ is the third expression

    $$\therefore P(x)=\left ( x^{2}+y^{2}+z^{2} \right )-\left ( x^{2}-y^{2}+5z^{2} \right )$$

    $$=x^{2}+y^{2}+z^{2}-x^{2}+y^{2}-5z^{2}=2y^{2}-4z^{2}$$
  • Question 4
    1 / -0
    $$\left( 4p{ x }^{ 2 }+5{ q }^{ 2 }y-9rz \right) -\left( -3{ q }^{ 2 }y+7p{ x }^{ 2 }-rz \right) =$$
    Solution
    $$(4px^{2}+5q^{2}y-9rz)-(-3q^{2}y+7px^{2}-rz)$$
    $$=4px^{2}+5q^{2}y-9rz+3q^{2}y-7px^{2}+rz$$
    $$=4px^{2}-7px^{2}+5q^{2}y+3q^{2}y-9rz+rz$$
    $$=-3px^{2}+8q^{2}y-8rz$$
  • Question 5
    1 / -0
    The sum of three expressions is $$x^{2}+y^{2}+z^{2}.$$ If two of them are $$4x^{2}-5y^{2}+z^{2}$$ and $$-3x^{2}+4y^{2}+2z^{2}$$, find the third expression.
    Solution
    Let the third expression be $$a$$ then,
    $$ a+(4x^{2}-5y^{2}+z^{2})+(-3x^{2}+4y^{2}+2z^{2}) = x^{2}+y^{2}+z^{2}$$
    $$ \Rightarrow a+(x^{2}-y^{2}+3z^{2})=(x^{2}+y^{2}+z^{2})$$
    $$ \Rightarrow a = (x^{2}+y^{2}+z^{2})-x^{2}+y^{2}-3z^{2}$$
    $$ \Rightarrow  a = 2y^{2}-2z^{2}$$

    Hence, option $$D$$ is correct.
  • Question 6
    1 / -0
    Subtract $$2x^{3}-4x^{2}+3x+5$$ from $$4x^{2}+x^{2}+x+6$$, then the resultant value is 
    Solution

  • Question 7
    1 / -0
    Add: $$2u+3v,\,\,-2v+3w,\,\,3u-v+2w$$
    Solution
    given, $$\left(2u+3v+3u-v+2w\right)+\left(-2v+3w\right)$$

    $$=5u+2v+2w-2v+3w$$

    $$=5u+5w$$
  • Question 8
    1 / -0
    Simplify: $$a-[b-\{c-(a-b-c)\}-c]+a$$
    Solution

  • Question 9
    1 / -0
    The sum of $$(6a+4b-c+3), (2b-3c+4), (11b-7a+2c-1)$$ and $$(2c-5a-6)$$ is
    Solution
    $$(6a+4b−c+3)+(2b−3c+4)+(11b−7a+2c−1)+(2c−5a−6)$$

    $$\Rightarrow$$  $$6a+4b−c+3+2b−3c+4+11b−7a+2c−1+2c−5a−6$$

    Now arranging and combining the like terms,

    $$\Rightarrow$$  $$(6a-7a-5a)+(4b+2b+11b)+(-c-3c+2c+2c)+(3+4-1-6)$$

    $$\Rightarrow$$  $$-6a+17b$$

    $$\therefore$$  The sum of $$(6a+4b−c+3),(2b−3c+4),(11b−7a+2c−1)$$ and $$(2c−5a−6)$$ is $$(-6a+17b)$$
  • Question 10
    1 / -0
    The sum of $${x}^{4}-xy+2{y}^{2}$$ and $$-{x}^{4}+xy+2{y}^{2}$$ is
    Solution
    Required sum $$={x}^{4}-xy+2{y}^{2}+-{x}^{4}+xy+2{y}^{2}$$ 
    $$={x}^{4}-xy+2{y}^{2}-{x}^{4}+xy+2{y}^{2}$$
    $$=[{x}^{4}+)-{x}^{4})]+(-xy+xy)+(2{y}^{2}+2{y}^{2})$$
    $$=0+0+4{y}^{2}=4{y}^{2}$$
    $$4{y}^{2}$$ is a monomial and polynomial in $$y$$
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