## How to Evaluate Two Variables?

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, $$4x^3$$ 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 $$x^3$$) and even in mixed form (like $$5x^2$$).
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 $$2(x^2 \ + \ y^3 \ – \ 25)$$
$$2(x^2) \ + \ 2(y^3) \ – \ 2(25) \ = \ 2(1^2) \ + \ 2(2^3) \ – \ 50 \ = \ 2 \ + \ 16 \ - \ 50 \ = \ -32$$.

### 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

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