Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Let's solve the equation [tex]\(\frac{1}{x+1} + \frac{1}{x+2} = \frac{2}{x+10}\)[/tex] step by step.
1. Identify the common denominator:
To add and equate the fractions, we first identify the common denominator for the terms on the left-hand side ([tex]\(\frac{1}{x+1}\)[/tex] and [tex]\(\frac{1}{x+2}\)[/tex]) and the right-hand side ([tex]\(\frac{2}{x+10}\)[/tex]) of the equation.
The common denominator for the left-hand side is [tex]\((x + 1)(x + 2)\)[/tex].
The common denominator for the right-hand side is not necessary to solve directly but it can be rewritten for comprehension.
2. Rewrite each term with the common denominator:
Let's rewrite each fraction over the common denominator:
[tex]\[ \frac{1}{x+1} = \frac{x+2}{(x+1)(x+2)} \][/tex]
[tex]\[ \frac{1}{x+2} = \frac{x+1}{(x+1)(x+2)} \][/tex]
3. Combine the fractions on the left-hand side:
Adding the fractions together:
[tex]\[ \frac{x+2}{(x+1)(x+2)} + \frac{x+1}{(x+1)(x+2)} = \frac{(x+2) + (x+1)}{(x+1)(x+2)} = \frac{2x + 3}{(x+1)(x+2)} \][/tex]
4. Equate the expression to the right-hand side:
Now, we will equate this to the right-hand side fraction:
[tex]\[ \frac{2x + 3}{(x+1)(x+2)} = \frac{2}{x+10} \][/tex]
5. Cross-multiply to solve for [tex]\( x \)[/tex]:
By cross-multiplying, we get:
[tex]\[ (2x + 3)(x + 10) = 2(x + 1)(x + 2) \][/tex]
6. Expand both sides of the equation:
[tex]\(\text{Left-hand side:}\)[/tex]
[tex]\[ (2x + 3)(x + 10) = 2x(x + 10) + 3(x + 10) = 2x^2 + 20x + 3x + 30 = 2x^2 + 23x + 30 \][/tex]
[tex]\(\text{Right-hand side:}\)[/tex]
[tex]\[ 2(x + 1)(x + 2) = 2(x^2 + 2x + 1x + 2) = 2(x^2 + 3x + 2) = 2x^2 + 6x + 4 \][/tex]
7. Set the expanded forms equal to each other:
Equate both simplified expressions:
[tex]\[ 2x^2 + 23x + 30 = 2x^2 + 6x + 4 \][/tex]
8. Solve the resulting linear equation:
Subtract [tex]\(2x^2\)[/tex] from both sides:
[tex]\[ 23x + 30 = 6x + 4 \][/tex]
Subtract [tex]\(6x\)[/tex] from both sides:
[tex]\[ 17x + 30 = 4 \][/tex]
Subtract 30 from both sides:
[tex]\[ 17x = -26 \][/tex]
Divide both sides by 17:
[tex]\[ x = -\frac{26}{17} \][/tex]
So, the solution to the equation [tex]\(\frac{1}{x+1} + \frac{1}{x+2} = \frac{2}{x+10}\)[/tex] is [tex]\(x = -\frac{26}{17}\)[/tex].
1. Identify the common denominator:
To add and equate the fractions, we first identify the common denominator for the terms on the left-hand side ([tex]\(\frac{1}{x+1}\)[/tex] and [tex]\(\frac{1}{x+2}\)[/tex]) and the right-hand side ([tex]\(\frac{2}{x+10}\)[/tex]) of the equation.
The common denominator for the left-hand side is [tex]\((x + 1)(x + 2)\)[/tex].
The common denominator for the right-hand side is not necessary to solve directly but it can be rewritten for comprehension.
2. Rewrite each term with the common denominator:
Let's rewrite each fraction over the common denominator:
[tex]\[ \frac{1}{x+1} = \frac{x+2}{(x+1)(x+2)} \][/tex]
[tex]\[ \frac{1}{x+2} = \frac{x+1}{(x+1)(x+2)} \][/tex]
3. Combine the fractions on the left-hand side:
Adding the fractions together:
[tex]\[ \frac{x+2}{(x+1)(x+2)} + \frac{x+1}{(x+1)(x+2)} = \frac{(x+2) + (x+1)}{(x+1)(x+2)} = \frac{2x + 3}{(x+1)(x+2)} \][/tex]
4. Equate the expression to the right-hand side:
Now, we will equate this to the right-hand side fraction:
[tex]\[ \frac{2x + 3}{(x+1)(x+2)} = \frac{2}{x+10} \][/tex]
5. Cross-multiply to solve for [tex]\( x \)[/tex]:
By cross-multiplying, we get:
[tex]\[ (2x + 3)(x + 10) = 2(x + 1)(x + 2) \][/tex]
6. Expand both sides of the equation:
[tex]\(\text{Left-hand side:}\)[/tex]
[tex]\[ (2x + 3)(x + 10) = 2x(x + 10) + 3(x + 10) = 2x^2 + 20x + 3x + 30 = 2x^2 + 23x + 30 \][/tex]
[tex]\(\text{Right-hand side:}\)[/tex]
[tex]\[ 2(x + 1)(x + 2) = 2(x^2 + 2x + 1x + 2) = 2(x^2 + 3x + 2) = 2x^2 + 6x + 4 \][/tex]
7. Set the expanded forms equal to each other:
Equate both simplified expressions:
[tex]\[ 2x^2 + 23x + 30 = 2x^2 + 6x + 4 \][/tex]
8. Solve the resulting linear equation:
Subtract [tex]\(2x^2\)[/tex] from both sides:
[tex]\[ 23x + 30 = 6x + 4 \][/tex]
Subtract [tex]\(6x\)[/tex] from both sides:
[tex]\[ 17x + 30 = 4 \][/tex]
Subtract 30 from both sides:
[tex]\[ 17x = -26 \][/tex]
Divide both sides by 17:
[tex]\[ x = -\frac{26}{17} \][/tex]
So, the solution to the equation [tex]\(\frac{1}{x+1} + \frac{1}{x+2} = \frac{2}{x+10}\)[/tex] is [tex]\(x = -\frac{26}{17}\)[/tex].
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
