Question

please answer it as soon as part​

Answer 1

hey mate,

(9z+7)-5(2z-3)+3(2z+3)

9z+7-10z+15+6z+9

9z-10z+6z+7+15+9

-z+6z+22+9

5z+31=(i)

I hope it helps u

Answer 2

Answer:

$\left(9z+7\right)-5\left(2z-3\right)+3\left(2z+3\right)=5z+31$

Step-by-step explanation:

$\left(9z+7\right)-5\left(2z-3\right)+3\left(2z+3\right)$

$\gray{\mathrm{Remove\:parentheses}:\quad \left(a\right)=a}$

$=9z+7-5\left(2z-3\right)+3\left(2z+3\right)$

$\black{\mathrm{Expand}\:-5\left(2z-3\right)}$

$-5\left(2z-3\right)$

$\gray{\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac}$

$\gray{a=-5,\:b=2z,\:c=3}$

$=-5\cdot \:2z-\left(-5\right)\cdot \:3$

$\gray{\mathrm{Apply\:minus-plus\:rules}}$

$\gray{a=-5,\:b=2z,\:c=3}$

$=-5\cdot \:2z+5\cdot \:3$

$\gray{\mathrm{Simplify}\:-5\cdot \:2z+5\cdot \:3:\quad -10z+15}$

$=-10z+15$

$=9z+7-10z+15+3\left(2z+3\right)$

$\black{\mathrm{Expand}\:3\left(2z+3\right)}$

$3\left(2z+3\right)$

$\gray{\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac}$

$\gray{a=3,\:b=2z,\:c=3}$

$=3\cdot \:2z+3\cdot \:3$

$\gray{\mathrm{Simplify}\:3\cdot \:2z+3\cdot \:3:\quad 6z+9}$

$=6z+9$

$=9z+7-10z+15+6z+9$

$\black{\mathrm{Simplify}\:9z+7-10z+15+6z+9}$

$9z+7-10z+15+6z+9$

$\gray{\mathrm{Group\:like\:terms}}$

$=9z-10z+6z+7+15+9$

$\gray{\mathrm{Add\:similar\:elements:}\:9z-10z+6z=5z}$

$=5z+7+15+9$

$\gray{\mathrm{Add\:the\:numbers:}\:7+15+9=31}$

$=5z+31$

$\therefore\left(9z+7\right)-5\left(2z-3\right)+3\left(2z+3\right)=5z+31\textrm{ [Option (i)]}$