Solve the rational equation. Check your solutions. (3)/(x+1)+(2)/(x-3)=4

Solve the rational equation. Check your solutions....

Key Concepts

Least Common Denominator (LCD)

The least common denominator is the smallest expression that can serve as a common denominator for all fractions involved. Multiplying both sides of the equation by the LCD eliminates the denominators and transforms the rational equation into a simpler, non-fractional equation.

Rational Equations

Rational equations are equations in which the expressions involve ratios of polynomials. They require careful manipulation because the variable may appear in the denominator, necessitating special techniques to combine terms and solve for the variable.

Excluded Values

When solving rational equations, it is important to identify and exclude any values that cause division by zero in the original equation. These values, known as excluded values, must be checked to ensure that the final solutions do not make any denominator zero.

Clearing Fractions

Clearing fractions involves multiplying each term of the equation by the LCD to remove the denominators, which simplifies the equation and allows for standard algebraic techniques, such as combining like terms and isolating the variable, to solve the equation.

Checking Solutions

After finding potential solutions by solving the cleared equation, it is essential to substitute them back into the original equation. This step verifies that the obtained solutions are valid and do not result in undefined expressions, such as division by zero in any of the original fractions.

Transcript

00:01 So we want to solve this equation and to do it, we have to work with the least common multiple of the denominator.

00:09 And it's going to be, once we have one as a denominator of four, we have that the least common multiple is x plus one times x minus three times one.

00:22 Okay, so we just have to multiply it by each term of the equation.

00:27 So here, x plus 1 times x minus 3 times 3 over x plus 1, okay, plus 2.

00:59 Let me put this common multiple first.

01:03 X plus 1 times x minus 3 times 2 over x minus 3 equal to.

01:19 4, okay? and here we equal to 4 times the list common multiple.

01:26 There is 4 times x plus 1 times x minus 3, okay? is this.

01:39 So now we can just cancel, cancel, cancel, cancel it.

01:49 And it's going to be three times this term, there is equal to 3x minus 9.

01:58 3x minus 9 plus 2 times this term that is going to be 2x plus 2.

02:10 Okay, it's equal to.

02:13 Here i'm just going to multiply this both terms here.

02:19 And it's going to be x times 4 times x times x squared minus 3x, x times x squared minus 3x, x times, x times, x squared, minus 3x, 1 times x plus 1 plus x, okay, 1 times minus 3.

02:41 There is minus 3.

02:46 So here we just have to multiply the 4, okay? and here we have 3x plus 2x and minus 9 plus 2 equal to 4x 4x squared.

03:15 Minus 12x plus 4x minus 12.

03:26 Okay? and here we are going to have 5x minus 7 equal to 4x squared minus 8x minus 12, okay? and the quadratic equation is going to be 4 x squared minus 13 x minus 5 equal to 0.

04:15 Okay, and we just have to solve this.

04:17 And we are going to work with the quadratic formula.

04:21 And here we have that delta is 13 squared minus 13 squared.

04:34 Okay, minus 4 times 4 times minus 5.

04:44 So the delta is going to be 249.

04:55 So we can now calculate the x, the solutions.

05:01 So x is going to be minus 13.

05:05 There is 13, okay, plus minus square root of 249.

05:19 Over 2 times 4.

05:24 So x1 is going to be equal to 13 plus square root of 249 over 8.

05:44 Okay.

05:45 And x2 is equal to 13 minus square root of 249 over 8.

06:02 Here is one.

06:06 So, the solutions are here, and we just have to substitute it in the equation, okay, to make sure that it is right.

06:22 So i'm going to substitute first the x -1, okay? and x -1 is this value, and substituting it, we have 3 over 13.

06:44 Plus square root of 249 plus 1 over 8 this plus 2 over x minus 1 so here i'm going to have x minus 3 okay if we look here it's x minus 3 so we're going to have 13 plus square root of 249 minus 3, okay, this over 8, it's equal to 4.

07:40 Okay? so we are just going to operate with this.

07:46 And here, we can do the least common multiple, the least common multiple, and this is going to be 3 over 13 plus square root of 249.

08:06 Plus one, this over eight...

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