Math / Algebra

QuestionQUESTION 7
What is the sum of all the solutions of the equation $\frac{x-1}{x+1}+\frac{4}{x-1}=\frac{5}{2}$ ?
A 0 C $\frac{4}{3}$ B $\frac{2}{3}$ D 2 a) b) c) d)

Studdy Solution

STEP 1

What is this asking? We need to find all the values of $x$ that make this equation true, and then add them together! Watch out! Remember to check if any solutions make the denominator zero, because that's a no-no!

STEP 2

1. Rewrite the equation
2. Solve for $x$
3. Check for invalid solutions
4. Calculate the sum

STEP 3

Let's make this equation easier to work with.
We can do this by getting rid of those pesky fractions!
Notice that $(x+1)$ and $(x-1)$ are in the denominators.
So, we can multiply both sides of the equation by $2 \cdot (x+1) \cdot (x-1)$ to get rid of the fractions.
Remember, we're multiplying every term by this!

STEP 4

$2 \cdot (x+1) \cdot (x-1) \cdot \frac{x-1}{x+1} + 2 \cdot (x+1) \cdot (x-1) \cdot \frac{4}{x-1} = 2 \cdot (x+1) \cdot (x-1) \cdot \frac{5}{2}$ Now, we can divide to one!

STEP 5

$2 \cdot (x-1) \cdot (x-1) + 2 \cdot (x+1) \cdot 4 = (x+1) \cdot (x-1) \cdot 5$ $2(x-1)^2 + 8(x+1) = 5(x+1)(x-1)$

STEP 6

Expand everything! $2(x^2 - 2x + 1) + 8x + 8 = 5(x^2 - 1)$ $2x^2 - 4x + 2 + 8x + 8 = 5x^2 - 5$

STEP 7

Combine like terms and move everything to one side to get a nice quadratic equation: $2x^2 + 4x + 10 = 5x^2 - 5$ $0 = 3x^2 - 4x - 15$

STEP 8

We can use the quadratic formula to solve for $x$ !
Remember, for an equation of the form $ax^2 + bx + c = 0$ , the solutions are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ In our case, $a = \mathbf{3}$ , $b = \mathbf{-4}$ , and $c = \mathbf{-15}$ .

STEP 9

Let's plug those values in: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-15)}}{2 \cdot 3}$ $x = \frac{4 \pm \sqrt{16 + 180}}{6}$ $x = \frac{4 \pm \sqrt{196}}{6}$ $x = \frac{4 \pm 14}{6}$

STEP 10

So, we have two possible solutions: $x = \frac{4 + 14}{6} = \frac{18}{6} = 3$ $x = \frac{4 - 14}{6} = \frac{-10}{6} = -\frac{5}{3}$

STEP 11

Remember, we can't have $x$ values that make the denominator zero in the original equation!
Let's check: * If $x = 3$ , the denominators are $3+1=4$ and $3-1=2$ .
These are fine! * If $x = -\frac{5}{3}$ , the denominators are $-\frac{5}{3} + 1 = -\frac{2}{3}$ and $-\frac{5}{3} - 1 = -\frac{8}{3}$ .
These are also fine!

STEP 12

Now, we just add our solutions together: $3 + \left(-\frac{5}{3}\right) = 3 - \frac{5}{3} = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}$

STEP 13

The sum of the solutions is $\frac{4}{3}$ .
So the answer is C!

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QuestionQUESTION 7
What is the sum of all the solutions of the equation $\frac{x-1}{x+1}+\frac{4}{x-1}=\frac{5}{2}$ ?
A 0 C $\frac{4}{3}$ B $\frac{2}{3}$ D 2 a) b) c) d)

Studdy Solution

STEP 1

What is this asking? We need to find all the values of $x$ that make this equation true, and then add them together! Watch out! Remember to check if any solutions make the denominator zero, because that's a no-no!

STEP 2

1. Rewrite the equation
2. Solve for $x$
3. Check for invalid solutions
4. Calculate the sum

STEP 3

STEP 4

$2 \cdot (x+1) \cdot (x-1) \cdot \frac{x-1}{x+1} + 2 \cdot (x+1) \cdot (x-1) \cdot \frac{4}{x-1} = 2 \cdot (x+1) \cdot (x-1) \cdot \frac{5}{2}$ Now, we can divide to one!

STEP 5

$2 \cdot (x-1) \cdot (x-1) + 2 \cdot (x+1) \cdot 4 = (x+1) \cdot (x-1) \cdot 5$ $2(x-1)^2 + 8(x+1) = 5(x+1)(x-1)$

STEP 6

Expand everything! $2(x^2 - 2x + 1) + 8x + 8 = 5(x^2 - 1)$ $2x^2 - 4x + 2 + 8x + 8 = 5x^2 - 5$

STEP 7

Combine like terms and move everything to one side to get a nice quadratic equation: $2x^2 + 4x + 10 = 5x^2 - 5$ $0 = 3x^2 - 4x - 15$

STEP 8

We can use the quadratic formula to solve for $x$ !
Remember, for an equation of the form $ax^2 + bx + c = 0$ , the solutions are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ In our case, $a = \mathbf{3}$ , $b = \mathbf{-4}$ , and $c = \mathbf{-15}$ .

STEP 9

Let's plug those values in: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-15)}}{2 \cdot 3}$ $x = \frac{4 \pm \sqrt{16 + 180}}{6}$ $x = \frac{4 \pm \sqrt{196}}{6}$ $x = \frac{4 \pm 14}{6}$

STEP 10

So, we have two possible solutions: $x = \frac{4 + 14}{6} = \frac{18}{6} = 3$ $x = \frac{4 - 14}{6} = \frac{-10}{6} = -\frac{5}{3}$

STEP 11

STEP 12

Now, we just add our solutions together: $3 + \left(-\frac{5}{3}\right) = 3 - \frac{5}{3} = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}$

STEP 13

The sum of the solutions is $\frac{4}{3}$ .
So the answer is C!

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QuestionQUESTION 7What is the sum of all the solutions of the equation x−1x+1+4x−1=52\frac{x-1}{x+1}+\frac{4}{x-1}=\frac{5}{2}x+1x−1​+x−14​=25​ ?A 0 C 43\frac{4}{3}34​ B 23\frac{2}{3}32​ D 2 a) b) c) d)

Studdy Solution

STEP 1

What is this asking? We need to find all the values of xxx that make this equation true, and then add them together! Watch out! Remember to check if any solutions make the denominator zero, because that's a no-no!

STEP 2

1. Rewrite the equation2. Solve for xxx3. Check for invalid solutions4. Calculate the sum

STEP 3

STEP 4

STEP 5

2⋅(x−1)⋅(x−1)+2⋅(x+1)⋅4=(x+1)⋅(x−1)⋅5 2 \cdot (x-1) \cdot (x-1) + 2 \cdot (x+1) \cdot 4 = (x+1) \cdot (x-1) \cdot 5 2⋅(x−1)⋅(x−1)+2⋅(x+1)⋅4=(x+1)⋅(x−1)⋅5 2(x−1)2+8(x+1)=5(x+1)(x−1) 2(x-1)^2 + 8(x+1) = 5(x+1)(x-1) 2(x−1)2+8(x+1)=5(x+1)(x−1)

STEP 6

Expand everything! 2(x2−2x+1)+8x+8=5(x2−1) 2(x^2 - 2x + 1) + 8x + 8 = 5(x^2 - 1) 2(x2−2x+1)+8x+8=5(x2−1) 2x2−4x+2+8x+8=5x2−5 2x^2 - 4x + 2 + 8x + 8 = 5x^2 - 5 2x2−4x+2+8x+8=5x2−5

STEP 7

Combine like terms and move everything to one side to get a nice quadratic equation: 2x2+4x+10=5x2−5 2x^2 + 4x + 10 = 5x^2 - 5 2x2+4x+10=5x2−5 0=3x2−4x−15 0 = 3x^2 - 4x - 15 0=3x2−4x−15

STEP 8

STEP 9

STEP 10

So, we have two possible solutions: x=4+146=186=3 x = \frac{4 + 14}{6} = \frac{18}{6} = 3 x=64+14​=618​=3 x=4−146=−106=−53 x = \frac{4 - 14}{6} = \frac{-10}{6} = -\frac{5}{3} x=64−14​=6−10​=−35​

STEP 11

STEP 12

Now, we just add our solutions together: 3+(−53)=3−53=93−53=43 3 + \left(-\frac{5}{3}\right) = 3 - \frac{5}{3} = \frac{9}{3} - \frac{5}{3} = \frac{4}{3} 3+(−35​)=3−35​=39​−35​=34​

STEP 13

The sum of the solutions is 43\frac{4}{3}34​.So the answer is C!

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QuestionQUESTION 7
What is the sum of all the solutions of the equation $\frac{x-1}{x+1}+\frac{4}{x-1}=\frac{5}{2}$ ?
A 0 C $\frac{4}{3}$ B $\frac{2}{3}$ D 2 a) b) c) d)

What is this asking? We need to find all the values of $x$ that make this equation true, and then add them together! Watch out! Remember to check if any solutions make the denominator zero, because that's a no-no!

1. Rewrite the equation
2. Solve for $x$
3. Check for invalid solutions
4. Calculate the sum

$2 \cdot (x-1) \cdot (x-1) + 2 \cdot (x+1) \cdot 4 = (x+1) \cdot (x-1) \cdot 5$ $2(x-1)^2 + 8(x+1) = 5(x+1)(x-1)$

Expand everything! $2(x^2 - 2x + 1) + 8x + 8 = 5(x^2 - 1)$ $2x^2 - 4x + 2 + 8x + 8 = 5x^2 - 5$

Combine like terms and move everything to one side to get a nice quadratic equation: $2x^2 + 4x + 10 = 5x^2 - 5$ $0 = 3x^2 - 4x - 15$

So, we have two possible solutions: $x = \frac{4 + 14}{6} = \frac{18}{6} = 3$ $x = \frac{4 - 14}{6} = \frac{-10}{6} = -\frac{5}{3}$

Now, we just add our solutions together: $3 + \left(-\frac{5}{3}\right) = 3 - \frac{5}{3} = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}$

The sum of the solutions is $\frac{4}{3}$ .
So the answer is C!