Question

$4a-6b=9$ , $b+4=8a$
Consider the system of equations above. Which of the following statement is true?

• A. There is only one solution $(a,b)\&a+b$ is positive
• B. There is only one solution $(a,b)\&a+b$ is negative
• C. There are infinitely many solutions
• D. There are no solutions

Answer 1

The correct option is B There is only one solution $(a,b)\&a+b$ is negative

Lets solve the equations using elimination method. Check ratio of coefficient of $x$ , ratio of coefficient of $y$ . As they are not equal, therefore option $C\&D$ are eliminated.
$4a-6b=9$
$-8a+b=4$
We multiply second equation by 6 .
$4a-6b=9$
$-48a+6b=24$
Adding the above two equations
$-44a=-15$
$a=\frac{15}{44}$
Next, lets substitute the values of $a$ into first equation and solve for $b$ :
$4\left(\frac{15}{44}\right)-6b=9$
$-6b=9-\frac{15}{11}$
$-6b=\frac{99-15}{11}$
$-6b=\frac{84}{11}$
$b=-\frac{84}{66}$
$b=-\frac{42}{33}$
Therefore, there is exactly one solution, $a+b=\frac{15}{44}-\frac{42}{33}$
$=\frac{45-168}{132}=-\frac{123}{132}<0$
Therefore correct answer option is B .