# Multiply: [(3/4)x - (4/3)y], [(2/3)x + (3/2)y]

**Solution:**

Given, __multiply__ [(3/4)x - (4/3)y], [(2/3)x + (3/2)y]

⇒ [(3/4)x - (4/3)y] × [(2/3)x + (3/2)y]

= [(1/2)x² + (9/8)xy - (8/9)xy - 2y²]

= (1/2)x² + (17/72)xy - 2y²

**✦ Try This:** Multiply: [(5/3)x - (2/3)y], [(3/5)x + (3/2)y]

Given, Multiply: [(5/3)x - (2/3)y], [(3/5)x + (3/2)y]

⇒ [(5/3)x - (2/3)y] × [(3/5)x + (3/2)y]

= x² + (5/2)xy - (2/5)xy - y²

= x² + (21/10)xy - y²

**☛ Also Check: **NCERT Solutions for Class 8 Maths Chapter 9

**NCERT Exemplar Class 8 Maths Chapter 7 Problem 83(xviii)**

## Multiply: [(3/4)x - (4/3)y], [(2/3)x + (3/2)y]

**Summary:**

Multiplying [(3/4)x - (4/3)y], [(2/3)x + (3/2)y] we get, (1/2)x² + (17/72)xy - 2y²

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