Q: 7r plus 3 plus 2r - 2?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

r+3+2r=21 to solver this equation you have to first simplify then solve. you have to combine same variables. 3r+3=21 -3 -3 you subtract three to get r by itself (3r=18)/3 you divide by three to get r by itself Answer r=6

-2r(3r - 11)(3r + 11)

0.3(2r) - 0.3(3) = 0.2(r) + 0.9Clear parentheses:0.6r - 0.9 = 0.2r + 0.9Multiply each side by 10:6r - 9 = 2r + 9Add 9 to each side:6r = 2r + 18Subtract 2r from each side:4r = 18Divide each side by 4:r = 18/4 = 4.5

That depends on what the question is. What you gave is simply a relationship between two variables. You can simplify it like this: 4x + 8 = 2r + 20 (4x + 8) / 2 = (2r + 20) / 2 2x + 4 = r + 10 2x + 4 - 4 = r + 10 - 4 2x = r + 6 You can then solve for x by dividing both sides by two: 2x / 2 = (r + 6) / 2 x = r/2 + 3 You could also solve for r by subtracting six from both sides, and rearranging the equation: 2x = r + 6 2x - 6 = r + 6 - 6 2x - 6 = r r = 2x - 6 r = 2(x - 3)

43

Related questions

12 + 8r - 6 = 7r - 5 - 3 8r +6 = 7r -8 r = -14

r+2

r+3+2r=21 to solver this equation you have to first simplify then solve. you have to combine same variables. 3r+3=21 -3 -3 you subtract three to get r by itself (3r=18)/3 you divide by three to get r by itself Answer r=6

-2r(3r - 11)(3r + 11)

If: 7 -2r = 13 Then: r = -3

0.3(2r) - 0.3(3) = 0.2(r) + 0.9Clear parentheses:0.6r - 0.9 = 0.2r + 0.9Multiply each side by 10:6r - 9 = 2r + 9Add 9 to each side:6r = 2r + 18Subtract 2r from each side:4r = 18Divide each side by 4:r = 18/4 = 4.5

Only if all the variables match up. Example of when they do match: 3x^2 +2y+r plus x^2- 2r+ 3y When they don't match: 3x^4+x+y plus 3x^3-x^2 +r

1)5r+7=13+2r+3r 2)5r+7=13+5r 3)5r's cancel 7=13 4)Subtract thirteen on both sides 7 =13 -13 -13 5)Answer is no solution because -6 doesn't equal to 0. -6=0 NO SOLUTION!!! Have fun with math!

Below ^ denotes power. 8q^6r^3=(2q^2r)^3 denote as a^3 and find a=2q^2r 27s^6t^3=(3s^2t)^3 denote as b^3 and find b=3s^2t Now 8q^6r^3+27s^6t^3=a^3+b^3 = (a+b)(a^2-ab+b^2) | substitute back =(2q^2r+3s^2t)(4q^4r^2-6q^2rs^2t+9s^4t^2). Notice (2q^2r)^2=4q^4r^2 (3s^2t)^2=9s^4t^2. Hence 8q^6r^3+27s^6t^3=(2rq^2+3ts^2)(4r^2q^4-6rts^2t^2+9t^2s^4), a=2q^2r and b=3s^2t.

That depends on what the question is. What you gave is simply a relationship between two variables. You can simplify it like this: 4x + 8 = 2r + 20 (4x + 8) / 2 = (2r + 20) / 2 2x + 4 = r + 10 2x + 4 - 4 = r + 10 - 4 2x = r + 6 You can then solve for x by dividing both sides by two: 2x / 2 = (r + 6) / 2 x = r/2 + 3 You could also solve for r by subtracting six from both sides, and rearranging the equation: 2x = r + 6 2x - 6 = r + 6 - 6 2x - 6 = r r = 2x - 6 r = 2(x - 3)

43

3 + 3 + 3 + 2 = 11