Removing Brackets Question....?

Question

Anybody got an answer for 5 (x-10) +3 (x+3)? And also tell me how you figured it out. I am moving into Year 8 and have Level 7 Work :D

Answer 1

5(x-10) + 3(x+3)

1.) distribute the number in front of the bracket to the numbers inside of it.

basically that means

5(x)-5(10) +3(x)+3(3)

you then get:

5x-50+3x+9

2) combine your like terms.

(5x+3x) and (-50+9)

so you have

8x - 41

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Answer 2

5x - 50 + 3x + 9. so its 8x - 41

Answer 3

the question is 5(x-10) + 3 (x+3)

so 5(x-10) can be expanded as 5*x - 5*10 = 5x - 50

similarly 3(x+3) = 3*x + 3*3 = 3x + 9

so the question now becomes

5x-50+3x+9

that adds up to 8x-41

[   if the equation is 5(x-10) + 3(x+3) = 0

then  8x-41 =0 which gives u an answer x= 41/8 ]

Answer 4

first of all you expand the formula.

1.5x-50+3x+9 if you don't know how to expand you take the number outside the bracket and times it with each letter or number.

2.8x-41 Now this is the tricky bit. You add the same numbers with the same numbers so 5x+3x is 8x. Now since its -50+9 the answer will be -41 because a two different symbols means you take away it which means getting it towards 0.

Answer 5

5(x-10) + 3(x+3)

= 5x - 50 + 3x + 9

= 5x + 3x - 50 +9

= 8x - 41

now, move the 41 to the other side by adding 41 to the right side

0 + 41 = 8x - 41 + 41

41 = 8x

41/8 = 8x/8

41/8 = x (you can also turn this to a fraction).