Solving For E And Evaluating Algebraic Expressions

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In this mathematical exploration, we delve into the realm of algebraic equations and expression evaluation. Our primary focus is to determine the value of the variable 'e' within the equation e=1x−e2y{ e = \frac{1}{x} - \frac{e}{2y} }, given specific values for 'x' and 'y'. Furthermore, we will embark on the evaluation of expressions involving variables 'g', 'h', and 'r', substituting their respective values to arrive at numerical solutions. This comprehensive approach will not only enhance our understanding of algebraic manipulations but also solidify our grasp of expression evaluation techniques.

Solving for 'e' in the Equation

To solve for 'e' in the equation e=1x−e2y{ e = \frac{1}{x} - \frac{e}{2y} }, we are given the values x=2{ x = 2 } and y=−3{ y = -3 }. The first step involves substituting these values into the equation. This substitution transforms the equation into a form where 'e' is the sole unknown variable, allowing us to isolate and determine its value. By replacing 'x' and 'y' with their numerical counterparts, we embark on a journey of algebraic manipulation to unravel the mystery surrounding 'e'.

Substituting the Values

Substituting x=2{ x = 2 } and y=−3{ y = -3 } into the equation e=1x−e2y{ e = \frac{1}{x} - \frac{e}{2y} }, we get:

e=12−e2(−3){ e = \frac{1}{2} - \frac{e}{2(-3)} }

This substitution is a pivotal step, as it transforms the equation from a general form involving variables to a specific instance with numerical values. The equation now contains only one unknown, 'e', making it solvable through algebraic manipulation. The next step involves simplifying the equation by performing the arithmetic operations and rearranging terms to isolate 'e' on one side.

Simplifying the Equation

Simplifying the equation further, we have:

e=12+e6{ e = \frac{1}{2} + \frac{e}{6} }

To solve for 'e', we need to eliminate the fraction involving 'e'. We can achieve this by subtracting e6{ \frac{e}{6} } from both sides of the equation. This operation maintains the equality while bringing all terms containing 'e' to one side, paving the way for isolating 'e'. The subsequent steps will involve combining like terms and performing further algebraic manipulations to arrive at the final solution.

Isolating 'e'

Subtracting e6{ \frac{e}{6} } from both sides:

e−e6=12{ e - \frac{e}{6} = \frac{1}{2} }

Now, we combine the terms on the left side. To do this, we need a common denominator, which in this case is 6. Rewriting 'e' as 6e6{ \frac{6e}{6} }, we get:

6e6−e6=12{ \frac{6e}{6} - \frac{e}{6} = \frac{1}{2} }

Combining the fractions on the left side simplifies the equation, bringing us closer to isolating 'e'. The next step involves performing the subtraction and then multiplying both sides of the equation by the reciprocal of the coefficient of 'e' to finally solve for its value.

Combining Terms

Combining the terms on the left side gives:

5e6=12{ \frac{5e}{6} = \frac{1}{2} }

To isolate 'e', we multiply both sides of the equation by the reciprocal of 56{ \frac{5}{6} }, which is 65{ \frac{6}{5} }. This operation will effectively cancel out the coefficient of 'e', leaving us with 'e' on one side and a numerical value on the other.

Solving for 'e'

Multiplying both sides by 65{ \frac{6}{5} }:

e=12×65{ e = \frac{1}{2} \times \frac{6}{5} }

e=610{ e = \frac{6}{10} }

Simplifying the fraction, we get:

e=35{ e = \frac{3}{5} }

Therefore, the value of 'e' in the equation e=1x−e2y{ e = \frac{1}{x} - \frac{e}{2y} }, given x=2{ x = 2 } and y=−3{ y = -3 }, is 35{ \frac{3}{5} } or 0.6. This completes the first part of our mathematical journey, where we successfully navigated the intricacies of algebraic manipulation to determine the value of an unknown variable.

Evaluating Expressions with Given Variables

Now, let's transition to the second part of our exploration: evaluating expressions. We are tasked with evaluating two expressions given the values g=0.2{ g = 0.2 }, h=−1{ h = -1 }, and r=3{ r = 3 }. This involves substituting these values into the expressions and performing the necessary arithmetic operations to arrive at numerical results. Expression evaluation is a fundamental skill in mathematics, allowing us to determine the value of complex combinations of variables and constants.

(a) Evaluating 5g2−(2h)3−4r{ 5g^2 - (2h)^3 - 4r }

Our first expression is 5g2−(2h)3−4r{ 5g^2 - (2h)^3 - 4r }. To evaluate this, we substitute the given values of g{ g }, h{ h }, and r{ r } into the expression. This substitution transforms the expression into a numerical calculation, which we can then simplify using the order of operations (PEMDAS/BODMAS). The key to accurate evaluation lies in careful substitution and adherence to the order of operations.

Substituting the Values

Substituting g=0.2{ g = 0.2 }, h=−1{ h = -1 }, and r=3{ r = 3 } into the expression, we get:

5(0.2)2−(2(−1))3−4(3){ 5(0.2)^2 - (2(-1))^3 - 4(3) }

This substitution replaces the variables with their numerical counterparts, setting the stage for the arithmetic operations. The next step involves simplifying the expression according to the order of operations, starting with exponents and then proceeding with multiplication, subtraction, and addition.

Simplifying the Expression

Simplifying the expression step by step:

First, we evaluate the exponents:

5(0.04)−(−2)3−4(3){ 5(0.04) - (-2)^3 - 4(3) }

5(0.04)−(−8)−4(3){ 5(0.04) - (-8) - 4(3) }

Next, we perform the multiplications:

0.2+8−12{ 0.2 + 8 - 12 }

Finally, we perform the addition and subtraction from left to right:

8.2−12{ 8.2 - 12 }

−3.8{ -3.8 }

Therefore, the value of the expression 5g2−(2h)3−4r{ 5g^2 - (2h)^3 - 4r } when g=0.2{ g = 0.2 }, h=−1{ h = -1 }, and r=3{ r = 3 } is −3.8{ -3.8 }.

(b) Evaluating the Second Expression (Expression Missing)

Unfortunately, the second expression to be evaluated is missing from the provided information. To complete this part of the task, we would need the expression involving 'g', 'h', and 'r'. Once the expression is provided, we can follow a similar process as in part (a): substitute the given values of g=0.2{ g = 0.2 }, h=−1{ h = -1 }, and r=3{ r = 3 } into the expression, and then simplify using the order of operations to obtain the numerical result.

Conclusion

In conclusion, we have successfully solved for 'e' in the equation e=1x−e2y{ e = \frac{1}{x} - \frac{e}{2y} }, given x=2{ x = 2 } and y=−3{ y = -3 }, finding its value to be 35{ \frac{3}{5} } or 0.6. We also evaluated the expression 5g2−(2h)3−4r{ 5g^2 - (2h)^3 - 4r } for g=0.2{ g = 0.2 }, h=−1{ h = -1 }, and r=3{ r = 3 }, obtaining a value of −3.8{ -3.8 }. The process involved substituting the given values into the expressions and simplifying using the order of operations. The second expression evaluation could not be completed due to the missing expression. This exercise has reinforced our understanding of algebraic manipulations and expression evaluation techniques, crucial skills in the realm of mathematics.

Repair Input Keyword: Given the formula e=1x−e2y{ e = \frac{1}{x} - \frac{e}{2y} } with x=2{ x = 2 } and y=−3{ y = -3 }, what is the value of e{ e }? Also, evaluate the expressions:

(a) 5g2−(2h)3−4r{ 5g^2 - (2h)^3 - 4r }

(b) (The expression is missing, please provide it)

when g=0.2{ g = 0.2 }, h=−1{ h = -1 }, and r=3{ r = 3 }.