Question

Please answer the number 14 & 15 ☺️

Answer 1

$a=-\dfrac{29}{60},\ b=2\dfrac{1}{3},\ c=-1\dfrac{3}{4}\\\\-\dfrac{1}{2}d=a+b+c\\\\\text{substitute}\\\\-\dfrac{1}{2}d=-\dfrac{29}{60}+2\dfrac{1}{3}+\left(-1\dfrac{3}{4}\right)\\\\-\dfrac{1}{2}d=-\dfrac{29}{60}+2\dfrac{1\cdot20}{3\cdot20}-1\dfrac{3\cdot15}{4\cdot15}\\\\-\dfrac{1}{2}d=-\dfrac{29}{60}+2\dfrac{20}{60}-1\dfrac{45}{60}\\\\-\dfrac{1}{2}d=\dfrac{120+20}{60}-\left(\dfrac{29}{60}+\dfrac{60+45}{60}\right)\\\\-\dfrac{1}{2}d=\dfrac{140}{60}-\dfrac{29+105}{60}$

$-\dfrac{1}{2}d=\dfrac{140}{60}-\dfrac{134}{60}\\\\-\dfrac{1}{2}d=\dfrac{6}{60}\\\\-\dfrac{1}{2}d=\dfrac{1}{10}\ \ \ \ |\cdot(-2)\\\\d=-\dfrac{1}{5}$