First solution :
Say, x = 0
→ 2x + 3y = 4
→ 2(0) + 3y = 4
→ 0 + 3y = 4
→ 3y = 4
→ y = 4/3
Second solution :
Say, x = 1
→ 2(1) + 3y = 4
→ 2 + 3y = 4
→ 3y = 4 - 2
→ 3y = 2
→ y = 2/3
Third solution :
Say, x = 2
→ 2(2) + 3y = 4
→ 4 + 3y = 4
→ 3y = 4 - 4
→ 3y = 0
→ y = 0 ÷ 3
→ y = 0
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