Chapter 1 - Introduction to Algebraic Expressions - 1.7 Multiplication and Division of Real Numbers - 1.7 Exercise Set - Page 59: 145

Step 1: Use Reverse of Distributive law as follows: $a(-b)+ab=a[-b+b]$ Step 2: Since the sum of the opposites is zero. We get, $a(-b)+ab=a(0)$ Step 3: Since the product of $0$ with any real number is $0$. We get, $a(-b)+ab=0$ Step 4: Since the sum of $a(-b)$ and $ab$ is zero. $a(-b)$ and $ab$ must be opposites of each other. Also, $a>0$ and $b>0$, that is, $ab>0$ Hence, $a(-b)$ is negative of $ab$.

Step 1: Use Reverse of Distributive law as follows: $a(-b)+ab=a[-b+b]$ Step 2: Since the sum of the opposites is zero. We get, $a(-b)+ab=a(0)$ Step 3: Since the product of $0$ with any real number is $0$. We get, $a(-b)+ab=0$ Step 4: Since the sum of $a(-b)$ and $ab$ is zero. $a(-b)$ and $ab$ must be opposites of each other. Also, $a>0$ and $b>0$, that is, $ab>0$ Hence, $a(-b)$ is negative of $ab$.