Equations with Algebraic Fractions

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply both sides by the nonzero value x: 8=2x.
2. Divide by 2 to get x=4.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply by x-1: 5=x-1.
2. Adding 1 gives x=6, which is allowed.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply both sides by x: x+3=2x.
2. Subtract x to get x=3.
3. This is nonzero, so it is a valid solution.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply both sides by (x+2)(x-1).
2. This gives 4(x-1)=2(x+2).
3. Expand: 4x-4=2x+4, so 2x=8.
4. Thus x=4. This does not make x+2 or x-1 zero, so it is the solution.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply by x+3: 2x-1=3(x+3).
2. Expand and collect terms: 2x-1=3x+9.
3. Hence x=-10, which is allowed.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Combine the fractions to get \frac{5}{x}=5.
2. Multiply by x to give 5=5x.
3. So x=1, which is nonzero.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply every term by x-2.
2. This gives x+4=2(x-2)+3.
3. Expand the right side: x+4=2x-1.
4. Therefore x=5, which is allowed.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply every term by the common denominator (x-1)(x+1) to clear the fractions.
2. This gives (x+1)+(x-1)=(x-1)(x+1), so 2x=x^2-1.
3. Rearrange into a quadratic equal to zero: x^2-2x-1=0.
4. This does not factorise, so use the quadratic formula with a=1, b=-2, c=-1: x=\frac{2\pm\sqrt{4+4}}{2}=\frac{2\pm\sqrt{8}}{2}.
5. \sqrt{8}=2\sqrt{2}, so x=1+\sqrt{2} or x=1-\sqrt{2}. Neither is 1 or -1, so both are solutions.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply by x to get 6=x(x-1).
2. Rearrange: x^2-x-6=0.
3. Factorise: (x-3)(x+2)=0.
4. The solutions are x=-2 and x=3, both nonzero.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply both sides by (x-2)(x+2).
2. This gives (x+1)(x+2)=(x-1)(x-2).
3. Expand: x^2+3x+2=x^2-3x+2.
4. The squared terms cancel, leaving 6x=0, so x=0.
5. The value 0 is allowed.

Factorise the numerator before deciding whether clearing the denominator is necessary.

1. Factorise the numerator as (x-3)(x+3).
2. x cannot be 3, as that would make the denominator zero, so cancel x-3, leaving x+3=8.
3. Thus x=5, which satisfies the original equation.

A value found after cancellation must still be allowed in the original equation.

1. Factorise the numerator: x^2+x-6=(x-2)(x+3).
2. For the allowed values, cancellation gives x+3=5.
3. This would require x=2, but x=2 makes the denominator x-2 zero, so it is not allowed in the original equation.
4. Therefore there are no solutions.

Multiply every term by x(x+2), then rearrange into a quadratic equal to zero.

1. Multiply every term by the common denominator x(x+2) to clear the fractions.
2. This gives 3(x+2)+4x=x(x+2).
3. Expand: 3x+6+4x=x^2+2x, so 7x+6=x^2+2x.
4. Rearrange into a quadratic equal to zero: x^2-5x-6=0.
5. Factorise: (x-6)(x+1)=0, so x=6 or x=-1.
6. Neither value makes a denominator zero, so both are solutions.

After solving, use the condition that x is positive.

1. Multiply by x+1: 4=(x-1)(x+1).
2. Thus x^2-1=4, so x^2=5.
3. The algebraic roots are \pm\sqrt{5}.
4. Only the positive root is allowed, so x=\sqrt{5}.

Write an equation using the number and its reciprocal, then clear the denominator.

1. Let the number be x>0. Its reciprocal is \frac{1}{x}.
2. Form the equation x+\frac{1}{x}=\frac{5}{2}.
3. Multiply through by 2x: 2x^2+2=5x.
4. Factorise the resulting quadratic: (2x-1)(x-2)=0.
5. The possible values are \frac{1}{2} and 2, and both are positive.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply every term by (x-1)(x+1): 2(x+1)-(x-1)=(x-1)(x+1).
2. Expand: 2x+2-x+1=x^2-1, so x+3=x^2-1.
3. Rearrange: x^2-x-4=0. This does not factorise.
4. Use the quadratic formula with a=1, b=-1, c=-4: x=\frac{1\pm\sqrt{1+16}}{2}=\frac{1\pm\sqrt{17}}{2}.
5. Neither value is 1 or -1, so x=\frac{1+\sqrt{17}}{2} or x=\frac{1-\sqrt{17}}{2}.

Clear both denominators, rearrange into a quadratic equal to zero, then use the quadratic formula.

1. Multiply every term by (x+2)(2x-1): 3(2x-1)+(x+2)=(x+2)(2x-1).
2. Expand both sides: 6x-3+x+2=2x^2-x+4x-2, so 7x-1=2x^2+3x-2.
3. Rearrange into a quadratic equal to zero: 2x^2-4x-1=0.
4. This does not factorise, so use the quadratic formula with a=2, b=-4, c=-1: x=\frac{4\pm\sqrt{16+8}}{4}=\frac{4\pm\sqrt{24}}{4}.
5. x=\frac{4+\sqrt{24}}{4}=2.2247\ldots or x=\frac{4-\sqrt{24}}{4}=-0.2247\ldots
6. Neither is -2 or \frac{1}{2}, so both are solutions. To 2 decimal places, x=2.22 or x=-0.22.

Write each journey time as distance \div speed, then use the difference of 1 hour to form an equation.

1. Time = distance \div speed, so the journey to the meeting takes \frac{120}{x} hours and the journey home takes \frac{120}{x+20} hours.
2. The journey home is 1 hour shorter: \frac{120}{x}-\frac{120}{x+20}=1.
3. Multiply every term by x(x+20): 120(x+20)-120x=x(x+20).
4. The left side simplifies to 2400, so 2400=x^2+20x, which rearranges to x^2+20x-2400=0.
5. Factorise: (x+60)(x-40)=0, so x=-60 or x=40.
6. A speed cannot be negative, so x=40. Check: 120\div40=3 hours and 120\div60=2 hours, which differ by 1 hour.

Multiply every term by a common denominator, then check that your answer does not make any denominator zero.

1. Multiply every term by 3(x-2)(x+2), which clears all three fractions.
2. This gives 3(x+2)+3(x-2)=(x-2)(x+2).
3. Simplify: 6x=x^2-4, so x^2-6x-4=0.
4. Use the quadratic formula with a=1, b=-6, c=-4: x=\frac{6\pm\sqrt{36+16}}{2}=\frac{6\pm\sqrt{52}}{2}.
5. \sqrt{52}=2\sqrt{13}, so x=3+\sqrt{13} or x=3-\sqrt{13}. Neither is 2 or -2, so both are solutions.

Convert the completion times into rates of work before forming the equation.

1. In one hour the first machine does \frac{1}{x} of the job and the second does \frac{1}{x+3} of the job.
2. Together they do the whole job in 2 hours, so they do \frac{1}{2} of it each hour: \frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}.
3. Multiply every term by 2x(x+3): 2(x+3)+2x=x(x+3).
4. Expand: 4x+6=x^2+3x, so x^2-x-6=0.
5. Factorise: (x-3)(x+2)=0, so x=3 or x=-2.
6. A time cannot be negative, so x=3. Check: in one hour the machines do \frac{1}{3}+\frac{1}{6}=\frac{1}{2} of the job.