Remainder theorem and simultaneous equations example

 Question: When a polynomial $P(x)$ is divided by $(x+2)(x-4)$ the quotient is the polynomial $Q(x)$ and the remainder is $ax+b$.  Find $a,b$ if $P(-2)=3$ and $P(4)=2$.

Solution.

Using the factor theorem we get two equations for $a,b$ which we can solve to find $a,b$.

We have $$P(x)=(x+2)(x-4)Q(x)+ax+b$$

Therefore

 $P(-2)=3=-2a+b\implies -2a+b=3$  (1)

and

 $P(4)=2=4a+b\implies 4a+b=2$  (2)

Subtract (2)-(2) gives

$-1=6a\implies a=-\dfrac{1}{6}$

Then $b=3+2(-1/6)=2\dfrac{2}{3}=8/3$.

Conclude: $$a=-\dfrac{1}{6}\:\ \mbox{ and }\:\   b=\dfrac{8}{3}$$.

Remember to check my working and tell me if I made an error!! (reward!)

Solve the pair of simultaneous equations $\log(x + y) = 0$ ...(1) $\quad\quad2 \log x = \log (y - 1) $ ...(2)

Solve the pair of simultaneous equations 

    $\log(x + y) = 0$             ...(1) 

    $2 \log x = \log (y - 1) $    ...(2) 

SOLUTION: 

$\log (x+y)=0$ in index form becomes $x+y = 10^0$ and so 

                     $x+y=1$      (1)   

In (2) we see that the domain of $x$ and $y$ is restricted by $x>0$ and $y-1$>0 (since the log function has domain positive numbers only)

           so we must have $x>0$ and $y>1$. 

Now using log laws on (2) and converting to index form we have 

                 $\log (x^2) = \log(y-1)$ 

                $\log(x^2) - \log(y-1) = 0$ 

             $  \log( x^2/(y-1) )  = 0 $

              $x^2/(y-1) = 10^0$             (index form of previous line) 

             $x^2/(y-1) = 1 $

            $x^2 = y - 1 $

            $y = x^2 + 1 $    (2) 

So we now have the two simultaneous equations to solve 

    $x+y=1$             (1)   
    $y = x^2 + 1$     (2) 
(but keeping in mind that $x>0, y>1$ must hold also) 

so,              $x + x^2 +1 = 1$ 
                 
                  $x^2 + x = 0$ 

                 $x ( x + 1 )= 0 $

               so this gives $x = 0, -1$ 
             but we must have $x>0$ and both of these solutions fail that condition. 

             So, we conclude that there are NO SOLUTIONS! 

Solve these equations $\log_{10} (xy)=7$ (1) $\log_{10} \left({x\over y}\right)=1$ (2)

Solve these equations
$\log_{10} (xy)=7$     (1)
$\log_{10} \left({x\over y}\right)=1$       (2)

SOLUTION:

Solve these equations $\log (p-q+1)=0$ (1) and $\log (pq)+1=0$ (2)

Solve these equations
$\log_{10} (p-q+1)=0$     (1)
$\log_{10} (pq)+1=0$       (2)

SOLUTION:

Grove Prelim 3U EX 3.13 - Q20 - Simultaneous equations,