Mathematics
Grade6
Easy

Question

Simplify fraction numerator 2 over denominator 3 end fraction y to the power of 3 end exponent plus fraction numerator 1 over denominator 4 end fraction y to the power of 3 end exponent plus fraction numerator 4 over denominator 5 end fraction x

  1. fraction numerator 11 y to the power of 3 end exponent over denominator 12 end fraction plus fraction numerator 4 x over denominator 5 end fraction    
  2. fraction numerator 11 y to the power of 2 end exponent over denominator 12 end fraction    
  3. fraction numerator 4 x over denominator 5 end fraction    
  4. xy    

hintHint:

This is an algebraic expression. An algebraic expression consists of two components: a variable and a constant. A variable is a quantity that doesn’t have any fixed value. It takes value according to the conditions. It is denoted by the alphabet. A constant is a quantity with a fixed value. In this question, we are given an expression. We have perform the required operations and simply the expression.

The correct answer is: fraction numerator 11 y to the power of 3 end exponent over denominator 12 end fraction plus fraction numerator 4 x over denominator 5 end fraction


    The given expression is 2 over 3 y cubed space plus space 1 fourth y cubed space plus space 4 over 5 x
    There are two variables in the above expression. The variables are: ‘x’ and ‘y’. There are three terms in the above expression. We have to perform the operations between them and simplify the expression.
    If we see, first and second term contains the same variable raised to same power 3. Such terms are called “like terms.” And the terms with different variables are called “unlike terms.”
    The third term has different variable.
    For like terms, we can perform the operations between their coefficients by keeping the variable constant. It means we can treat each term as a constant and perform operations like normal operations on constants.
    For the above expression, we can perform an operation between the first term and second term as both the terms as they have the same variable ‘y3'. We will keep ‘y3’ constant and perform addition between the coefficients.
    Let’s see the expression
    2 over 3 y cubed space plus space 1 fourth y to the power of 3 space end exponent plus 4 over 5 x space equals space 8 over 12 y cubed space plus space 3 over 12 y to the power of 3 space end exponent plus 4 over 5 x
    To make the denominators of the first and second terms the same, we multiply and divide the first term by the denominator of the second term and vice versa.
    Solving further we get,
    2 over 3 y cubed space plus space 1 fourth y to the power of 3 space end exponent plus 4 over 5 x space equals space 11 over 12 y cubed space plus space 4 over 5 x
    So this is the final value.

    We have to be careful about the power of variable to. The variables with same power can operate. If we have same variable but different power no further operation is possible.

    Related Questions to study

    card img

    With Turito Academy.

    card img

    With Turito Foundation.

    card img

    Get an Expert Advice From Turito.

    Turito Academy

    card img

    With Turito Academy.

    Test Prep

    card img

    With Turito Foundation.