How do you solve #(7x) /(2x+5) + 1 = (10x - 3) /( 3x)#?

Question

Nathan Axel · Accepted Answer

#x=(29+-sqrt(421))/14#

The first thing you do is incorporate the 1 on the left side into the fraction. In order to do that you substitute it by a fraction that is the left denominator divided by itself, like this:

#(7x)/(2x+5)+(2x+5)/(2x+5)=(10x-3)/(3x)#

Now we can add the fractions on the left side. Since they have the same denominator, we only add the numerators.

#(7x+2x+5)/(2x+5)=(10x-3)/(3x)#

#(9x+5)/(2x+5)=(10x-3)/(3x)#

Now we multiply both sides of the equation to remove the #3x# from the bottom of the right side.

#(3x*(9x+5))/(2x+5)=(10x-3)/cancel(3x)*cancel(3x)#

Multiply the top on the left and we get:

#(3x*9x+3x*5)/(2x+5)=10x-3#

#(27x^2+15x)/(2x+5)=(10x-3)#

Do the same thing with the left denominator. Multiply both sides by it in order to remove it.

#cancel(2x+5)(27x^2+15x)/cancel(2x+5)=(10x-3)*(2x+5)#

Now we multiply out the right side:

#27x^2+15x=(10x-3)(2x+5)#

#27x^2+15x=20x^2+50x-6x-15#

#27x^2+15x=20x^2+44x-15#

Finally we can move everything from the right side of the equation to the left:

#27x^2+15x-20x^2-44x+15=0#

#7x^2-29x+15=0#

Now we have a single equation if the form of #ax^2+bx+c=0#

Which means we can use the quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)# to solve for #x#

When we plug in the numbers, we get:

#x=(29+-sqrt((-29)^2-4*7*15))/(2*7)#

#x=(29+-sqrt(841-420))/14#

#x=(29+-sqrt(421))/14#

Savannah Anderson · Answer

undefined To solve the equation (7x) /(2x+5) + 1 = (10x - 3) /( 3x), we can follow these steps:

1. Multiply both sides of the equation by the common denominator, which is (2x+5)(3x). This will eliminate the denominators.

(7x) /(2x+5) + 1 = (10x - 3) /( 3x)
[(7x)(2x+5)(3x)] /(2x+5) + (2x+5)(3x) = [(10x - 3)(2x+5)(3x)] /( 3x)

2. Simplify the equation by canceling out common factors.

7x(3x) + (2x+5)(3x)(2x+5) = (10x - 3)(2x+5)(3x)

3. Expand and simplify both sides of the equation.

21x^2 + 3x(2x+5)(2x+5) = (10x - 3)(2x+5)(3x)

4. Expand and simplify further.

21x^2 + 3x(4x^2 + 20x + 25) = (10x - 3)(6x^2 + 30x)

5. Distribute and simplify.

21x^2 + 12x^3 + 60x^2 + 75x = 60x^3 + 300x^2 - 18x^2 - 90x

6. Combine like terms.

21x^2 + 12x^3 + 60x^2 + 75x = 60x^3 + 300x^2 - 18x^2 - 90x

7. Rearrange the equation to bring all terms to one side.

12x^3 + 60x^2 + 21x^2 + 75x - 60x^3 - 300x^2 + 18x^2 + 90x = 0

8. Combine like terms.

-48x^3 - 204x^2 + 186x = 0

9. Factor out common terms.

-6x(8x^2 + 34x - 31) = 0

10. Set each factor equal to zero and solve for x.

-6x = 0 or 8x^2 + 34x - 31 = 0

11. Solve the first equation.

x = 0

12. Solve the second equation using factoring, quadratic formula, or completing the square.

x = (-34 ± √(34^2 - 4(8)(-31))) / (2(8))

x = (-34 ± √(1156 + 992)) / 16

x = (-34 ± √2148) / 16

x ≈ -3.07 or x ≈ 0.24

Therefore, the solutions to the equation are x = 0, x ≈ -3.07, and x ≈ 0.24.