Master Simple Equations with Step-by-Step Examples | Math Homework Help

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In summary, simple equations are mathematical expressions involving a single variable that can be solved by performing operations on both sides of the equation to isolate the variable. To solve a simple equation, you need to perform the same operation on both sides of the equation, such as addition, subtraction, multiplication, or division. An example of solving a simple equation is 5x + 10 = 30, where we subtract 10 from both sides and then divide by 5 to get x = 4. Some common mistakes when solving simple equations include forgetting to perform the same operation on both sides and not properly distributing or combining like terms. To check your answer, you can substitute the value you found for the variable back into the original equation to see if
  • #1
ImAnEngineer
209
1

Homework Statement


a=(4b-14)/(b-5)

b= ...

2. The attempt at a solution
Attempted a lot of things, but couldn't isolate the b. Is it possible at all to do this, when there are multiple terms in the numerator. If so, how?
 
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  • #2
Never mind, problem solved.

b= 5a-14 / a-4
 
  • #3


Yes, it is possible to solve for b in this equation. Here is a step-by-step example:

1. Distribute the 4 to the terms inside the parentheses:
a = (4b-14)/(b-5)
a = (4b-20+6)/(b-5)

2. Combine like terms inside the parentheses:
a = (4b-14)/(b-5)
a = (4b-14+6)/(b-5)
a = (4b-8)/(b-5)

3. Factor out a common factor of 4 in the numerator:
a = (4b-8)/(b-5)
a = 4(b-2)/(b-5)

4. Now we can see that the denominator is the difference of two terms, so we can use the difference of squares formula to simplify it:
a = 4(b-2)/((b-5)(b+5))

5. Now we can set this equal to b and solve for b:
b = 4(b-2)/((b-5)(b+5))

6. Multiply both sides by the denominator to get rid of the fraction:
b(b-5)(b+5) = 4(b-2)

7. Expand the left side using the distributive property:
b^3-5b^2+5b^2-25b = 4b-8

8. Combine like terms on both sides:
b^3-20b+8 = 0

9. Use the rational root theorem to find possible rational solutions:
Possible solutions: ±1, ±2, ±4, ±8

10. Use synthetic division to test the possible solutions:
b=-2 is a solution, so (b+2) is a factor of the polynomial.

11. Use synthetic division again to divide the polynomial by (b+2):
b^3-20b+8 = (b+2)(b^2-2b-4)

12. Use the quadratic formula to solve for the remaining solutions:
b^2-2b-4 = 0
b = (2±√(4+4(4)))/2
b = (2±√20)/2
b = 1±√5

Therefore, the solutions to the equation are b = -2, 1+√5, and 1-√5.
 

1. What are simple equations?

Simple equations are mathematical expressions that involve a single variable and can be solved by performing operations on both sides of the equation to isolate the variable.

2. How do I solve simple equations?

To solve a simple equation, you need to perform the same operation on both sides of the equation to isolate the variable. This can include addition, subtraction, multiplication, and division.

3. Can you provide an example of solving a simple equation?

Sure, let's say we have the equation 5x + 10 = 30. To solve for x, we need to get rid of the 10 on the left side. We can do this by subtracting 10 from both sides, giving us 5x = 20. Then, we can divide both sides by 5 to get x = 4. So the solution to this simple equation is x = 4.

4. What are some common mistakes when solving simple equations?

One common mistake is forgetting to perform the same operation on both sides of the equation. Another mistake is not properly distributing or combining like terms before solving.

5. How can I check my answer when solving a simple equation?

You can check your answer by substituting the value you found for the variable back into the original equation and seeing if it is true. For example, in our previous equation, we can check by substituting x = 4 to get 5(4) + 10 = 30, which is true.

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