Question

Solve in positive integers:
$6x+7y+4z=122$
$11x+8y-6z=145$.

Answer 1

$6x+7y+4z=122⟶1$

$11x+8y-6z=145⟶2$

Multiplying by 8 in equation 1 and by 7 in equation 2

$48x+56y+32z=976$

$77x+56y-42z=1015$ [signs will be changed]

$\overline{-29x+74z=-39}$

$let,z=a$

$-29x+74a=-39$

$29x=74a+39$

Now putting this value in equation1

$6\left(\frac{74a+39}{29}\right)+7y+4a=122$

$444a+234+203y+116a=3538$

$203y+560a=3304⟶3$

$11\left(\frac{74a+39}{29}\right)+8y-6a=145$

$814a+429+232y-174a=420$

$232y+640a=3776⟶4$

$203y+560a=3304$ [signs will be changed]

$\overline{29y+80a=472}$

$⟹y=\frac{472-80a}{29}$

So, the solutions are,

$\{\left(\frac{74a+39}{29}\right),\left(\frac{472-80a}{29}\right),a\}$ Now, putting $a=3,x=9,y=8,z=3$

$\therefore$So, the solution to positive integers is $x=9,y=8,z=3.$