Well, we can use some simple addition, okay? So if Ally has 7 cubes, and Greg has four, we can use the equation: x+y = ?, so 7 + 4 = 11. x and y don't mean anything in this scenario, but I just used them as variables. I hope this helps! :)

7 0

3 years ago

Read 2 more answers

Please do this math for me or else i will steal your cookies

kompoz [17]

Answer:

Step-by-step explanation:

$-7x+16=58\\\\\mathrm{Subtract\:}16\mathrm{\:from\:both\:sides}\\-7x+16-16=58-16\\\\-7x=42\\\\\mathrm{Divide\:both\:sides\:by\:}-7\\\\\frac{-7x}{-7}=\frac{42}{-7}\\\\x =-6$

$-2x+15=-9\\\\\mathrm{Subtract\:}15\mathrm{\:from\:both\:sides}\\\\-2x+15-15=-9-15\\\\-2x=-24\\\\\mathrm{Divide\:both\:sides\:by\:}-2\\\\\frac{-2x}{-2}=\frac{-24}{-2}\\\\x=12\\$

$5x-4=36\\\\\mathrm{Add\:}4\mathrm{\:to\:both\:sides}\\\\5x-4+4=36+4\\\\5x=40\\\\\mathrm{Divide\:both\:sides\:by\:}5\\\\\frac{5x}{5}=\frac{40}{5}\\\\x=8$

$25-3x=88\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\\\25-3x-25=88-25\\\\\mathrm{Divide\:both\:sides\:by\:}-3\\\\\frac{-3x}{-3}=\frac{63}{-3}\\\\x=-21$

$-11=7-x\\\\\mathrm{Add\:}x\mathrm{\:to\:both\:sides}\\\\-11+x=7-x+x\\\\-11+x=7\\\\\mathrm{Add\:}11\mathrm{\:to\:both\:sides}\\\\-11+x+11=7+11\\\\x=18$

$65+15x=35\\\\\mathrm{Subtract\:}65\mathrm{\:from\:both\:sides}\\\\65+15x-65=35-65\\\\\mathrm{Divide\:both\:sides\:by\:}15\\\\\frac{15x}{15}=\frac{-30}{15}\\\\x=-2$

$\frac{1}{2} x-18=2\\\\\mathrm{Add\:}18\mathrm{\:to\:both\:sides}\\\\\frac{1}{2}x-18+18=2+18\\\\\frac{1}{2}x=20\\\\\mathrm{Multiply\:both\:sides\:by\:}2\\\\2\times\frac{1}{2}x=20\times \:2\\\\x=4$

$\frac{2}{3}x-10=-12 \\\\\mathrm{Add\:}10\mathrm{\:to\:both\:sides}\\\\\frac{2}{3}x-10+10=-12+10\\\\\frac{2}{3}x=-2\\\\Divide\:both\:sides\:by\: 2/3\\\\\frac{2}{3}x\div \frac{2}{3} =-2\div \frac{2}{3} \\\\\frac{2}{3} x\times \frac{3}{2} =-2\times\frac{3}{2} \\\\x =-3$

$6-\frac{1}{3}x=-1 \\\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\\\6-\frac{1}{3}x-6=-1-6\\\\-\frac{1}{3}x=-7\\\\\mathrm{Multiply\:both\:sides\:by\:}-3\\\\\left(-\frac{1}{3}x\right)\left(-3\right)=\left(-7\right)\left(-3\right)\\\\x=21$

10.

$4-9x=-14\\\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\\\4-9x-4=-14-4\\\\-9x=-18\\\\\mathrm{Divide\:both\:sides\:by\:}-9\\\\\frac{-9x}{-9}=\frac{-18}{-9}\\\\x=2$

11.

$11-x=29\\\\\mathrm{Subtract\:}11\mathrm{\:from\:both\:sides}\\\\-x=18\\\\\mathrm{Divide\:both\:sides\:by\:}-1\\\\\frac{-x}{-1}=\frac{18}{-1}\\\\x=-18$

12.

$-9-11x=68\\\\\mathrm{Add\:}9\mathrm{\:to\:both\:sides}\\\\-9-11x+9=68+9\\\\-11x=77\\\\\mathrm{Divide\:both\:sides\:by\:}-11\\\\\frac{-11x}{-11}=\frac{77}{-11}\\\\x=-7$

13.

$45+\frac{5}{6}x =50\\\\\mathrm{Subtract\:}45\mathrm{\:from\:both\:sides}\\\\45+\frac{5}{6}x-45=50-45\\\\\frac{5}{6}x=5\\\\\mathrm{Divide\:both\:sides\:by\:}5/6\\\\\frac{5}{6} x\div\frac{5}{6}=5\div\frac{5}{6}\\\\\frac{5}{6}x\times \frac{6}{5}= 5\times\frac{6}{5}\\\\ x=6$

14.

$-5x+17=-33\\\\\mathrm{Subtract\:}17\mathrm{\:from\:both\:sides}\\\\-5x+17-17=-33-17\\\\\mathrm{Divide\:both\:sides\:by\:}-5\\\\\frac{-5x}{-5}=\frac{-50}{-5}\\\\x=10$

15.

$95=-4+33x\\\\-4+33x=95\\\\\mathrm{Add\:}4\mathrm{\:to\:both\:sides}\\\\-4+33x+4=95+4\\\\\mathrm{Divide\:both\:sides\:by\:}33\\\\\frac{33x}{33}=\frac{99}{33}\\\\x=3$