84 of 100: Adding is Multiplying is Dividing?

Make three systems of equations.

$A+B\quad =\quad AB\\ A+B\quad =\quad \frac { A }{ B } \\ AB\quad \quad =\quad \frac { A }{ B }$

Let's start with $AB\quad =\quad \frac { A }{ B } \\$

Solve B and we get 1 or -1.

If we plug in 1 for $A+B\quad=\quad AB$ , we get $A+1\quad =\quad A$ . And since we know that doesn't exist, The answer for B is -1.

Plug -1 for B in either $A+B\quad =\quad \frac { A }{ B }$ or $A+B=\quad AB$ and we get $A=\frac { 1 }{ 2 }$ .

The answer is $(\frac { 1 }{ 2 } ,-1)$ . So therefore it has only one ordered pair.

$\begin{aligned} a\times b & = \dfrac{a}{b} \\ \Longrightarrow a\times b^2 & =a\end{aligned}$ Note that $a\neq 0$ , since if $a=0$ , then $b=0$ , but then $\dfrac{a}{b}$ makes error. So we can divide by $a$ . $\begin{aligned} a\times b^2 & = a \\ b^2 & = 1 \\ & \begin{cases} b=1 \\ b=-1 \end{cases} \end{aligned}$

Now let's look the first two requirements. $\begin{aligned} a\times b & = a+b \\ \Longrightarrow ab-(a+b) & = 0 \\ ab-a-b+1 & = 1 \\ (a-1)(b-1)& =1 \end{aligned}$ $b$ can't be $1$ , because then $(a-1)(b-1)=0$ , so $b=-1$ . Then $\begin{aligned} (a-1)(b-1) & = 1 \\ (a-1)(-1-1) & = 1 \\ -2\times (a-1) & = 1 \\ a-1 & = -\dfrac{1}{2} \\ a & = \dfrac{1}{2} \end{aligned}$ This gives us the only solution.

Therefore there is $\boxed{\text{one}}$ ordered pair.

First, let's start by looking at ab = $\frac{a}{b}$ . Diving both sides by a, we get b = $\frac{1}{b}$ . Multiplying both sides by b results in b^2 = 1. Rooting both sides gives us b=1 or b=-1. Now let's look at a+b = ab. First we'll substitute 1 for b. This results in a+1 = a; However, this is't possible as using SPOE to subtract a from both sides results in 1 = 0, which isn't true. That leaves b = -1. Using that, we get a - 1 = -a. Adding a to both sides with APOE gives 2a -1 = 0, which leads to a = $\frac{1}{2}$ . We now have one ordered, and only one ordered pair, that works: ( $\frac{1}{2}$ , -1). Ans. one

The first equality may be written $(a-1)(b-1) = 1.$ This equation has infinitely many solutions: all values of $a$ are possible, except $a \not= 1$ . The same is true for $b$ .

The second equality is true whenever $a = 0$ . However, in that case the first equality gives $(-1)\cdot (b - 1) = 1$ so that $b = 0$ ; but this must be ruled out because we are not allowed to divide by zero.

The second equality is also true whenever $b^2 = 1$ , that is, $b = \pm 1$ . We already concluded that $b \not= 1$ . That leaves $\boxed{one}$ solution: $b = -1, a = \frac12:\ \ \ \ \ \ \ \ \frac12 + (-1) = \frac12\times(-1) = \frac{1/2}{-1} = -\frac12.$

The first equation a + b = ab, then we have a = -b/(1 – b), The second equation a + b = a/b, then a = b^2/(1 – b), The third equation ab = a/b, b^2= 1, b=1, or -1 (NA) The solution is only b=-1, a=1/2

First look at the equation $a\times b=\frac{a}b$ . Simplifying this, we get $b^2=1$ , so $b=\pm1$ . However, when plugging in $b=1$ , we get $a+1=a$ from the first two expressions, so $b$ should be $-1$ . Now, we solve the equation $a-1 = -a$ and get the single solution $a=0.5, b=-1$ .

From $a\times b = \dfrac ab$ , $\implies b^2 = 1$ $\implies b = \pm 1$ . Now consider $a+b = a\times b$ .

$\begin{cases} b = 1 & \implies a+1=a & & \color{#D61F06} \text{No solution} \\ b = -1 & \implies a-1=-a & \implies a = \dfrac 12 & \color{#3D99F6} \text{Only solution} \end{cases}$

Therefore there is only $\boxed{1}$ solution $\left(\frac 12, - 1\right)$ .

From $a + b = ab$ it follows that $\frac{a}{b} = a - 1 .\quad$

$a - 1$ must be the value for all three expressions or there is no solution.

If $a + b = a - 1$ then $b = -1$

Since $b = -1$ it follows that $ab = -a = a -1$ and $a = \frac{1}{2}$

Therefore there is just one ordered pair $(\frac{1}{2}, -1)$

For a*b to equal a/b, b must equal 1 or negative one. If b=1, then a+1=a, which is unsolvable. If b=-1, then a-1=-a, so 2a-1=0, so 2a=1, so a=1/2.

a + b = ab = a/b,
=> a = ab - b = b(a - 1),
=> a/b = a - 1 = a + b,
=> b = -1,
Now, a + b = ab, so a - 1 = -a,
Or, 2a = 1, so, a = 1/2 and b = -1 is the only solution.

It is $b\neq 0$ , since it is in the denominator.

From $a+b=\frac{a}{b}$ we get $a=ab+b^2$ .

Substituting in $ab=a+b$ yields $a=a+b+b^2$ , which is equivalent to $0=b+b^2=b(b+1)$ .

Since $b\neq 0$ , we may divide by b without losing a solution and get 0=b+1, i.e. b=-1.

So a-1=-a and a=0.5.

Hence, there is at most one solution. If we substitute a and b with 0.5 and -1, we see that this is indeed a solution.

a+b = a/b

ab + b ^2 = a

a(b-1) = -b^2

a= -b^2/b-1

ab = -b^3/b-1

a/b = -b/b-1

ab = a/b

-b^3/b-1 = -b/b-1 only If

b = -1 or b = 0

Let us examine b=-1

Then a=1/2

a+b = -1/2 = ab = a/b

Obviously a=-1 , b = ½ does not work since a/b = -2 which is not a+b = -1/2.

Let us examine b = 0

Then a+b = ab only if a=0 and b = 0

But a/b = 0/0 which is not equal to zero but undefined

So, the only ordered pair that works is (1/2 -1)

b*b=1

Therefore a=0.5

First, let's solve $ab = \frac{a}{b}$ :

$\displaystyle{ab = \frac{a}{b}} \\ \Rightarrow ab^2 = a \\ \Rightarrow a(b-1)(b+1) = 0$

This gives us three possibilities: $a = 0$ , $b = 1$ and $b = -1$ .

Now let's try to solve $a + b = ab$ for each case.

Substituting $a = 0$ gives $b = 0$ . But $b$ can't be 0, as $\frac{a}{b}$ is then undefined.

Substituting $b = 1$ gives $a + 1 = a$ , which is impossible.

Substituting $b = -1$ gives $a - 1 = -a \\ \Rightarrow \displaystyle{a = \frac12}$

Testing the combination, we see that $\displaystyle{\frac12 + (-1) = \frac12 \times (-1) = \frac{\frac12}{-1} = -\frac12}$ .

Thus $\displaystyle{\boxed{a = \frac12, b = -1}}$ is the only solution.