We can solve the equation by performing inverse operation until x is isolated. However, first we should add the fractions on the left-hand side by giving them the same denominator.

3/x+1+4/x=2

ExpandFrac

a/b=a * (x+1)/b * (x+1)

3/x+1+4(x+1)/x(x+1)=2

ExpandFrac

a/b=a * x/b * x

3x/x(x+1)+4(x+1)/x(x+1)=2

AddFrac

Add fractions

3x+4(x+1)/x(x+1)=2

MultEqn

LHS * x(x+1)=RHS* x(x+1)

3x+4(x+1)=2x(x+1)

Distr

Distribute 4 & 2x

3x+4x+4=2x^2+2x

AddTerms

Add terms

7x+4=2x^2+2x

SubEqn

LHS-7x=RHS-7x

4=2x^2-5x

SubEqn

LHS-4=RHS-4

0=2x^2-5x-4

We can solve this equation by using the Quadratic Formula.

0=2x^2-5x-4

UseQuadForm

Use the Quadratic Formula: a = 2, b= - 5, c= - 4

x=-( - 5) ± sqrt(( - 5)^2-4 * 2( - 4))/2 * 2

NegNeg

- (- a)=a

x=5 ± sqrt((- 5)^2-4 * 2(- 4))/2 * 2

CalcPowProd

Calculate power and product

x=5 ± sqrt(25+32)/4

AddTerms

Add terms

x=5 ± sqrt(57)/4

StateSol

State solutions

lx=.(4+sqrt(57)) /4. x=.(4-sqrt(57)) /4.

UseCalc

Use a calculator

lx=-0.63745... x=3.13745...

RoundDec

Round to 2 decimal place(s)

lx_1=-0.64 x_2=3.14

Let's try to solve the equation in the same way as in Parts A and B.

3/x+2+5=3/x+2

SubEqn

LHS-3/x+2=RHS-3/x+2

5≠ 0

The equation has no solution.

An equation states that what is on the left-hand side and on the right-hand side are equal. Examining the equation, we notice that the two sides have an identical term.

3/x+2+5= 3/x+2 If we add or subtract something to either side of the equation, we have to add or subtract the same thing to the other side. However, in the equation from Part C, we have only added 5 to the left-hand side. Therefore, this equation cannot have a solution, as the left-hand side will always be 5 greater than the right-hand side.