Question

which values of a and b make the following equation true?
$(5x^{5}y^{2})(-4x^{4}y^{8})=-20x^{a}y^{b}$
$a = 11, b = 7$
$a = 11, b = 10$
$a = 28, b = 7$
$a = 28, b = 10$

Explanation:

Step1: Multiply coefficients and variables separately

First, multiply the coefficients: \(5\times(-4)= -20\). Then, for the \(x\)-terms, use the rule of exponents \(x^m\times x^n = x^{m + n}\). So, \(x^7\times x^4=x^{7 + 4}=x^{11}\). For the \(y\)-terms, \(y^2\times y^5=y^{2+5}=y^7\). So, \((5x^7y^2)(-4x^4y^5)=-20x^{11}y^7\).

Step2: Compare with the given equation

The given equation is \(-20x^{11}y^7=-20x^a y^b\). By comparing the exponents of \(x\) and \(y\), we get \(a = 11\) and \(b = 7\).

Answer:

a = 11, b = 7 (corresponding to the first option: a = 11, b = 7)