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how do you solve 9c(c-11)+10(5c-3)=3c(c+5)+c(6c-3)-30 if we are multiplying polynomials by monomials

Sagot :

[tex]9c(c-11)+10(5c-3)=3c(c+5)+c(6c-3)-30\\\\9c(c)+9c(-11)+10(5c)+10(-3)=3c(c)+3c(5)+c(6c)+c(-3)-30\\\\9c^2-99c+50c-30=3c^2+15c+6c^2-3c-30\\\\9c^2-49c-30=9c^2+12c-30\ \ \ |subtract\ (9c^2-30)\ from\ both\ sides\\\\-49c=12c\ \ \ \ |subtract\ 12c\ from\ both\ sides\\\\-61c=0\ \ \ \ |divide\ both\ sides\ by\ (-61)\\\\\boxed{c=0}[/tex]
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