[tex] 3(x + 5) - 2(x + 2) = 20 \\
3x + 15 - 2x - 4 = 20\\
5x + 11 = 20\\
5x = 9\\
x = 1\frac{4}{5} [tex]
Answer:
15x^3y+6x^2y^2+9xy^3
Step-by-step explanation:
3xy ( 5x^2 +2xy +3y^2)
Distribute the 3xy to all terms in the parentheses
3xy*5x^2+3xy*2xy+3xy*3y^2
15x^3y+6x^2y^2+9xy^3
No If it was 4 to 3 then yes but 4 can't be multiplied by anything to equal 9
I dont completely know but it might be 3
