Solve {l}{7x+2/2x-3+5x+4/x=34x^2+43x-2/4x^2-9+10-x/2x^2-3x}{y=333}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { \frac{7 x + 2}{2 x - 3} + \frac{5 x + 4}{x} = \frac{34 x ^ {2} + 43 x - 2}{4 x ^ {2} - 9} + \frac{10 - x}{2 x ^ {2} - 3 x} }\\ { y = 333 }\\ { z = y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for x, y, z, a, b, c

$$c=333$$

Solution Steps

Consider the first equation. Variable x cannot be equal to any of the values -\frac{3}{2},0,\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by x\left(2x-3\right)\left(2x+3\right), the least common multiple of 2x-3,x,4x^{2}-9,2x^{2}-3x.

$$x\left(2x+3\right)\left(7x+2\right)+\left(4x^{2}-9\right)\left(5x+4\right)=x\left(34x^{2}+43x-2\right)+\left(2x+3\right)\left(10-x\right)$$

Use the distributive property to multiply x by 2x+3.

$$\left(2x^{2}+3x\right)\left(7x+2\right)+\left(4x^{2}-9\right)\left(5x+4\right)=x\left(34x^{2}+43x-2\right)+\left(2x+3\right)\left(10-x\right)$$

Use the distributive property to multiply 2x^{2}+3x by 7x+2 and combine like terms.

$$14x^{3}+25x^{2}+6x+\left(4x^{2}-9\right)\left(5x+4\right)=x\left(34x^{2}+43x-2\right)+\left(2x+3\right)\left(10-x\right)$$

Use the distributive property to multiply 4x^{2}-9 by 5x+4.

$$14x^{3}+25x^{2}+6x+20x^{3}+16x^{2}-45x-36=x\left(34x^{2}+43x-2\right)+\left(2x+3\right)\left(10-x\right)$$

Combine 14x^{3} and 20x^{3} to get 34x^{3}.

$$34x^{3}+25x^{2}+6x+16x^{2}-45x-36=x\left(34x^{2}+43x-2\right)+\left(2x+3\right)\left(10-x\right)$$

Combine 25x^{2} and 16x^{2} to get 41x^{2}.

$$34x^{3}+41x^{2}+6x-45x-36=x\left(34x^{2}+43x-2\right)+\left(2x+3\right)\left(10-x\right)$$

Combine 6x and -45x to get -39x.

$$34x^{3}+41x^{2}-39x-36=x\left(34x^{2}+43x-2\right)+\left(2x+3\right)\left(10-x\right)$$

Use the distributive property to multiply x by 34x^{2}+43x-2.

$$34x^{3}+41x^{2}-39x-36=34x^{3}+43x^{2}-2x+\left(2x+3\right)\left(10-x\right)$$

Use the distributive property to multiply 2x+3 by 10-x and combine like terms.

$$34x^{3}+41x^{2}-39x-36=34x^{3}+43x^{2}-2x+17x-2x^{2}+30$$

Combine -2x and 17x to get 15x.

$$34x^{3}+41x^{2}-39x-36=34x^{3}+43x^{2}+15x-2x^{2}+30$$

Combine 43x^{2} and -2x^{2} to get 41x^{2}.

$$34x^{3}+41x^{2}-39x-36=34x^{3}+41x^{2}+15x+30$$

Subtract 34x^{3} from both sides.

$$34x^{3}+41x^{2}-39x-36-34x^{3}=41x^{2}+15x+30$$

Combine 34x^{3} and -34x^{3} to get 0.

$$41x^{2}-39x-36=41x^{2}+15x+30$$

Subtract 41x^{2} from both sides.

$$41x^{2}-39x-36-41x^{2}=15x+30$$

Combine 41x^{2} and -41x^{2} to get 0.

$$-39x-36=15x+30$$

Subtract 15x from both sides.

$$-39x-36-15x=30$$

Combine -39x and -15x to get -54x.

$$-54x-36=30$$

Add 36 to both sides.

$$-54x=30+36$$

Add 30 and 36 to get 66.

$$-54x=66$$

Divide both sides by -54.

$$x=\frac{66}{-54}$$

Reduce the fraction \frac{66}{-54} to lowest terms by extracting and canceling out 6.

$$x=-\frac{11}{9}$$

The system is now solved.

$$x=-\frac{11}{9} y=333 z=333 a=333 b=333 c=333$$