How to Evaluate Two Variables

How to Evaluate Two Variables?

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So, in mathematics, we refer to a variable as an unknown entity or something that can have no fixed value. For example, let us consider the case of natural numbers. In a set of natural numbers, n can be defined as anything between 1 and infinity. So, n belongs to the set of Integers (1, 2, 3, 4, ………). Therefore, in this case we can consider n as a variable because it has some definite value, but that value is not fixed. It completely depends on the numerical problem that what value of a particular variable should we use.
Moreover, in an expression, a variable can have coefficients and exponential powers. For example, 4x3 is a variable with an exponent power of 3 and a co-efficient value of 4. So, variables can be found in just pure form (for example x3) and even in mixed form (like 5x2).
Also, in an expression, we can simplify the variable terms by grouping them into like and unlike terms. We should perform all mathematical operations separately to these like and unlike term groups.

How to Evaluate Two Variables?

To evaluate two variables, we must follow the given steps:

  • If possible, first simplify the variable expression
  • Next, just substitute the value of the variables in the equation.

Example Questions:

  • Substitute for x = 2, y = 6 in 5(x \ + \ y) \ – \ 6
    5x \ + \ 5y \ – \ 6 \ = \ 5(2) \ + \ 5(6) \ – \ 6 \ = \ 10 \ + \ 30 \ - \ 6 \ = \ 34.
  • Substitute for x \ = \ 3, \ y \ = \ 4 in 5(x \ + \ y \ – \ xy)
    5x \ + \ 5y \ – \ 5xy \ = \ 5(3) \ + \ 5(4) \ – \ 5(3)(4) \ = \ 15 \ + \ 20 \ - \ 60 \ = \ -25.
  • Substitute for x \ = \ 1, \ y \ = \ 2 in 2(x^2 \ + \ y^3 \ – \ 25)
    2(x^2) \ + \ 2(y^3) \ – \ 2(25) \ = \ 2(1^2) \ + \ 2(2^3) \ – \ 50 \ = \ 2 \ + \ 16 \ - \ 50 \ = \ -32.

Free printable Worksheets

Exercises for Evaluating Two Variables

1)  x \ = \ 10, y \ = \ 5, \ -5(4x \ - \ 2y \ - \ 10) =

2)  x \ = \ 17, y \ = \ 2, \ -2(2x \ - \ 3y \ - \ 17) =

3)  x \ = \ 4, y \ = \ 5, \ 2x \ + \ \frac{15}{y} \ - \ 4 =

4)  x \ = \ 2, y \ = \ 3, \ 4x \ + \ 6y \ - \ 2 =

5)  x \ = \ 8, y \ = \ 4, \ 4(3x \ - \ 3y \ + \ 8) =

6)  x \ = \ 9, y \ = \ 7, \ 4x \ + \ 36y \ - \ 9 =

7)  x \ = \ 6, y \ = \ 3, \ 2x(2y \ - \ 6) =

8)  x \ = \ 15, y \ = \ 4, \ 4(4x \ - \ 3y \ + \ 15) =

9)  x \ = \ 19, y \ = \ 4, \ 2x \ + \ \frac{12}{y} \ - \ 19 =

10)  x \ = \ 2, y \ = \ 2, \ -2(4x \ - \ 4y \ - \ 2) =

 
1) x \ = \ 10, y \ = \ 5, \ -5(4x \ - \ 2y \ - \ 10) = \ \color{red}{ -5(4 \times 10 \ - \ 2 \ \times \ 5 \ - \ 10) \ = \ -100}
2) x \ = \ 17, y \ = \ 2, \ -2(2x \ - \ 3y \ - \ 17) = \ \color{red}{ -2(2 \times 17 \ - \ 3 \ \times \ 2 \ - \ 17) \ = \ -22}
3) x \ = \ 4, y \ = \ 5, \ 2x \ + \ \frac{15}{y} \ - \ 4 = \ \color{red}{2 \times 4 \ + \ \frac{15}{5} \ - \ 4 \ = \ 7}
4) x \ = \ 2, y \ = \ 3, \ 4x \ + \ 6y \ - \ 2 = \ \color{red}{4 \times 2 \ + \ 6 \ \times \ 3 \ - \ 2 \ = \ 24}
5) x \ = \ 8, y \ = \ 4, \ 4(3x \ - \ 3y \ + \ 8) = \ \color{red}{ 4(3 \times 8 \ - \ 3 \ \times \ 4 \ + \ 8) \ = \ 80}
6) x \ = \ 9, y \ = \ 7, \ 4x \ + \ 36y \ - \ 9 = \ \color{red}{4 \times 9 \ + \ 36 \ \times \ 7 \ - \ 9 \ = \ 279}
7) x \ = \ 6, y \ = \ 3, \ 2x(2y \ - \ 6) = \ \color{red}{ (2 \times 6)(2 \times 3 \ - \ 6) \ = \ 0}
8) x \ = \ 15, y \ = \ 4, \ 4(4x \ - \ 3y \ + \ 15) = \ \color{red}{ 4(4 \times 15 \ - \ 3 \ \times \ 4 \ + \ 15) \ = \ 252}
9) x \ = \ 19, y \ = \ 4, \ 2x \ + \ \frac{12}{y} \ - \ 19 = \ \color{red}{2 \times 19 \ + \ \frac{12}{4} \ - \ 19 \ = \ 22}
10) x \ = \ 2, y \ = \ 2, \ -2(4x \ - \ 4y \ - \ 2) = \ \color{red}{ -2(4 \times 2 \ - \ 4 \ \times \ 2 \ - \ 2) \ = \ 4}

Evaluating Two Variables Practice Quiz