Mathematics 10 — Exercise 5.2 · Q1 (vii)

Full question

Simplify the following sum or difference of algebraic fractions.

Step-by-step solution

1. Fraction 1: keep its denominator nonzero and factor that denominator.

   $$x+y\ne 0$$

2. Fraction 2: keep its denominator nonzero and factor that denominator.

   $$x-y\ne 0$$

3. Fraction 3: keep its denominator nonzero and factor that denominator.

   $$x^2-y^2\ne 0$$

4. Expand any powers or inner brackets before factoring.

   $$x^2-y^2=x^2-y^2$$

5. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

   $$x^2-y^2=\left(x-y\right)\times \left(x+y\right)$$

6. The complete factored form is:

   $$\left(x-y\right)\times \left(x+y\right)$$

7. Build the least common denominator from every distinct denominator factor at its largest required power.

   $$\left(x-y\right)\times \left(x+y\right)$$

8. For fraction 1, divide the common denominator by its own denominator.

   $$\frac{x^2-y^2}{x+y}=x-y$$

9. Multiply this numerator by the same factor to keep the fraction unchanged.

   $$\frac{\left(2\times x\right)\times \left(x-y\right)}{\left(x-y\right)\times \left(x+y\right)}$$

10. For fraction 2, divide the common denominator by its own denominator.

    $$\frac{x^2-y^2}{x-y}=x+y$$

11. Multiply this numerator by the same factor to keep the fraction unchanged.

    $$\frac{\left(-y\right)\times \left(x+y\right)}{\left(x-y\right)\times \left(x+y\right)}$$

12. For fraction 3, divide the common denominator by its own denominator.

    $$\frac{x^2-y^2}{x^2-y^2}=1$$

13. Multiply this numerator by the same factor to keep the fraction unchanged.

    $$\frac{1\times \left(-2\times y^2\right)}{\left(x-y\right)\times \left(x+y\right)}$$

14. The denominators now match. Put all signed numerators over one common denominator.

    $$\frac{\left(2\times x\right)\times \left(x-y\right)+\left(-y\right)\times \left(x+y\right)+1\times \left(-2\times y^2\right)}{\left(x-y\right)\times \left(x+y\right)}$$

15. Expand numerator product 1, keeping minus signs on the whole bracket.

    $$\left(2\times x\right)\times \left(x-y\right)=2\times x^2-2\times x\times y$$

16. Expand numerator product 2, keeping minus signs on the whole bracket.

    $$\left(-y\right)\times \left(x+y\right)=-x\times y-y^2$$

17. Expand numerator product 3, keeping minus signs on the whole bracket.

    $$1\times \left(-2\times y^2\right)=-2\times y^2$$

18. Collect like powers and constants in the combined numerator.

    $$2\times x^2-2\times x\times y-(x\times y+y^2)-2\times y^2=2\times x^2-3\times x\times y-3\times y^2$$

19. The original denominator must be nonzero. Keep this restriction even if a factor is later cancelled.

    $$x^2-y^2\ne 0$$

20. Factor the numerator.

21. Factor the denominator.

22. Expand any powers or inner brackets before factoring.

    $$x^2-y^2=x^2-y^2$$

23. Apply the matching factor identity: difference of squares, sum/difference of cubes, or grouping. Expand the displayed factors to check the identity.

    $$x^2-y^2=\left(x-y\right)\times \left(x+y\right)$$

24. The complete factored form is:

    $$\left(x-y\right)\times \left(x+y\right)$$

25. Write the factored numerator over the factored denominator.

    $$\frac{2\times x^2-3\times x\times y-3\times y^2}{\left(x-y\right)\times \left(x+y\right)}$$

26. Write the fraction in lowest form. Only factors, never added terms, have been cancelled.

    $$\frac{2\times x^2-3\times x\times y-3\times y^2}{\left(x-y\right)\times \left(x+y\right)}$$

Answer

The final answer is shown below.