Solve each equation. Look A likes ⋯a. (1)/(x)=(2 x)/(x+1) b. (1)/(x)+(2 x)/(x+1)=1

Name: Problem 85, Chapter 10: Elementary and Intermediate Algebra
Uploaded: 2020-04-23T10:39:31-08:00
Duration: 6 min 10 s
Channel: Shaza Hammoud
Description: Solve each equation. Look A likes ⋯a. (1)/(x)=(2 x)/(x+1) b. (1)/(x)+(2 x)/(x+1)=1

Solve each equation. Look A likes ⋯a. (1)/(x)=(2 x)/(x+1) b....

Key Concepts

Clearing Denominators

Clearing denominators involves multiplying both sides of the equation by the least common denominator (LCD) to eliminate the fractions. This step simplifies the equation, converts it to a polynomial equation, and makes it easier to solve, while still requiring a check for any extraneous solutions.

Quadratic Equations

After clearing the denominators from a rational equation, the result can often be a quadratic equation. Solving quadratic equations typically involves methods such as factoring, completing the square, or using the quadratic formula. It is important to check the solutions against the domain restrictions to ensure all solutions are valid.

Rational Equations

Rational equations are equations that involve fractions containing polynomials in the numerator and the denominator. Solving these equations requires careful manipulation because the variable appears in the denominators, which means that some values may be excluded from the solution set due to resulting in division by zero.

Domain Restrictions

When working with rational equations, it is essential to identify and exclude values that make any denominator equal to zero. These restrictions prevent undefined expressions and guarantee that only valid solutions are considered in the final answer.

Transcript

00:01 This question asks to solve the following two parts, we have two equations, which they look the same, but they are small different where the part b, the two fractions are in the right left -hand side, and they are equal to one.

00:17 And here, one of the fractions is in the left -hand side, and the other one is the right -hand side.

00:23 So in this part, let's try to solve for x.

00:28 And since this fraction is in left -hand side and the other fraction is the right -hand side, let's use the cross -product.

00:35 So x will be multiplied by 2x and the result will appear here, and x plus 1 will be multiplied by 1, and the result will be all appear here.

00:46 So 1 multiplied by x plus 1 is x plus 1, so this is in the left -hand side, and the right -hand side will be 2x multiplied by x, and this is 2x, 2.

00:58 Now, let's shift the 2x to the other side to make all the terms in one side and the right answer will be equal to 0.

01:07 So this is minus 2x12 plus x plus 1 equals to 0.

01:18 Now what's needed to be done is to solve this quadratic equation using one of the methods.

01:26 I'll use the quadratic formula where a, b, and c can be.

01:29 Obtained directly from the equation.

01:32 So a is the coefficient of x squared, b is the coefficient of x, which is 1, and c is the constant number 1.

01:40 So this is x sub 1 and 2 equals to b minus 1, as it's minus p, so it's minus 1 plus minus the square root of b2 of 2, which is 1 to the power of 2, which is 1, minus 4a is minus 2 and c is 1 this is over 2a and a is minus 2 so this is minus 1 and under the square root we have 1 minus 4mata by minus 2 is 8 so 1 minus minus 8 will be positive so 1 plus 8 is 9 under the square root and this is plus minus 3 over minus 4.

02:29 So we have two solutions.

02:31 One is when taking the positive sign of 3, this is minus 1 plus 3, which is positive 2.

02:38 2 over minus 4 is minus 1 over 2.

02:42 And the other one is by taking the negative sign of 3, and this is minus 1, minus 1, minus 3 is minus 4, over minus 4 over minus 4 is 1.

02:49 So these are the two solutions of this equation.

02:53 So now let's solve the equation in part b...

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