Examples with solutions for Domain: Exercises with fractions

Exercise #1

6x+5=1 \frac{6}{x+5}=1

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To solve this problem, we will determine the domain, or field of application, of the equation 6x+5=1 \frac{6}{x+5} = 1 .

Step-by-step solution:

  • Step 1: Identify the denominator. In the given equation, the denominator is x+5 x+5 .
  • Step 2: Determine when the denominator is zero. Solve for x x by setting x+5=0 x+5 = 0 .
  • Step 3: Solve the equation: x+5=0 x+5 = 0 gives x=5 x = -5 .
  • Step 4: Exclude this value from the domain. The domain is all real numbers except x=5 x = -5 .

Therefore, the field of application of the equation is all real numbers except where x=5 x = -5 .

Thus, the domain is x5 x \neq -5 .

Answer

x5 x\operatorname{\ne}-5

Exercise #2

25a+4b7y+43+2=9b \frac{25a+4b}{7y+4\cdot3+2}=9b

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To solve the problem, follow these steps:

  • Step 1: Understand that the equation 25a+4b7y+43+2=9b\frac{25a+4b}{7y + 4 \cdot 3 + 2}=9b is undefined when the denominator equals zero.
  • Step 2: Simplify the denominator: 7y+43+27y + 4 \cdot 3 + 2.
  • Step 3: Calculate the constant part: 43=124 \cdot 3 = 12, so the expression becomes 7y+12+27y + 12 + 2.
  • Step 4: Combine constants: 12+2=1412 + 2 = 14. The denominator is 7y+147y + 14.
  • Step 5: Set the denominator equal to zero to find values of yy that make the equation undefined: 7y+14=07y + 14 = 0.
  • Step 6: Solve for yy:
    • Subtract 14 from both sides: 7y=147y = -14.
    • Divide by 7: y=2y = -2.

Therefore, the equation is undefined when y=2y = -2. The field of application excludes y=2y = -2.

The choice that reflects this is y2\boxed{y \neq -2}.

Answer

y2 y\operatorname{\ne}-2

Exercise #3

xyz2(3+y)+4=8 \frac{xyz}{2(3+y)+4}=8

What is the field of application of the equation?

Video Solution

Step-by-Step Solution

To find the domain of the given equation xyz2(3+y)+4=8 \frac{xyz}{2(3+y)+4}=8 , we need to ensure the denominator is not zero. This means solving 2(3+y)+4=0 2(3+y) + 4 = 0 .

Let's solve this step-by-step:

  • Step 1: Simplify the expression 2(3+y)+4=0 2(3+y) + 4 = 0 .
  • Step 2: Expand to 6+2y+4=0 6 + 2y + 4 = 0 .
  • Step 3: Combine like terms to get 2y+10=0 2y + 10 = 0 .
  • Step 4: Isolate the variable y y . Subtract 10 from both sides: 2y=10 2y = -10 .
  • Step 5: Divide by 2 to solve for y y : y=5 y = -5 .

If y=5 y = -5 , the denominator becomes zero, which makes the original expression undefined.

Therefore, the value of y y must not be 5-5 for the expression to be valid. In conclusion, the restriction on y y is that y5 y \neq -5 .

The correct answer choice is: y5 y \neq -5 .

Answer

y5 y\ne-5