algebra1

could you walk-through the steps to solving this problem because I am stuck;
3x-5(3x-7)=2(x+9)+45

Let me know if you have a question in any step.

3x-5(3x-7)=2(x+9)+45
3x-15x+35= 2x+18+45
-12x+35 = 2x+63
-35 -35
-12x= 2x +63-35
-12x= 2x +28
-2x -2x

-12x-2x = 28

-14x= 28
-14x/-14= 28/-14
x = -2

to check just substitute the answer for x to ever x in the original problem.

3x-5(3x-7)=2(x+9)+45

3(-2)-5(3(-2)-7)=2(-2+9)+45
-6 -5 (-6-7)= 2(7)+45
-6-5(-13)=14+45
-6+65=59
59=59 it checks.


You really helped me out. I know see how to work out other problems like this one. Thank you very much!!!

asked by unknown

Respond to this Question

First Name

Your Response

Similar Questions

  1. algbrea

    3x-5(3x-7)= 2(x+9) + 45 3x-15x+35= 2x+18 +45 -12x + 35 = 2x+63 -12x -28 = 2x -12 -28 =x -40 =x 3x-5(3x-7)= 2(x+9) + 45 3x-15x+35= 2x+18 +45 -12x + 35 = 2x+63 -10x+35=63 -10x=28 x=-2.8 you are both wrong! 3x - 15x + 35 = 2x + 18 +
  2. Algebra. Please check my answers

    Evaluate a/b+c^d for a=6,b=3,c=4 and d=5 A.42*** B.18 C.82 D.360 What is the simplified form of the expression ? -(12x-8y) A.12x+8y B.12x-8y***** C.-12x+8y D.-12x-8y
  3. Polynomial Fucntion Problem

    Could you help me with the following problem, I don't understand how to do it. Am I suppose to use the linear factorization theorem? Find an nth degree polynomial function with real coefficients satisfying the given conditions. 1.
  4. algebra

    24x^4-12x^3+48x^2/-4x 24x^4-12x^3+12(-1) 24x^4-12x^3+(-12x) 2x^4-12x^3-12x check this problem for me please
  5. algebra2

    1/3 - 5/6 = 1/X SOLVE 1/3 - 5/6 = 1/X Let's remove the denominators by multiplying by 12X 12X*(1/3) - 12X*(5/6) = 12X*(1/X) Clear the fractions. 4X - 10X = 12 -6X=12 multipy by -1 6X=-12 Divide both sides by 6 6X/6 = -12/6 X = -2
  6. math

    how do i do x2+12x+32 x2+12x = 14x+32 subtract 14x from each side 14x=32 divide by 2 7x=16 x=16/7, which is 2 and 2/7 ok maybe you typed that question wrong? is it x^2 + 12x + 32 ? if it is then the solution is: (x+8)(x+4) because
  7. calc

    What is dy/dx of y=12/12X-y^3 ???? Any idea anyone??? y=12/12X-y^3 I assume you mean... y=12/(12X-y^3) 12Xy-y^4=12 12x dy/dx + 12y -4y^3 dy/dx=0 dy/dx (12x -4y^3)=-12y solve for dy/dx
  8. Math

    Can you correct the errors? 4(x+3)=24 4(3x)=24 12x=24 12x/12= 24/12 X=2 5(x+2)=45 5x+10=45 15x=45 15x/15=45/15 x=3 5+4x=13 5+4x-4x=13-4x 5=9x 5/9=9x/9 5/9= x 56=8(x+3) 56=8x+24 56-24=8x+24-24 32=8x 32/32=8x/32 x=8/32 x=1/4
  9. Calculus!! HELP!!

    1) The sides of the triangle shown increase in such a way that (dz/dt=1) and (dx/dt=(3dy/dx)) At the instant when x = 12 and y = 5, what is the value of dx/dt? 2) Let f(x) = x^3 − 4. Which of these is the equation for the normal
  10. math

    3x^2+6x/2x^2-11x+12 - 2x^2-5x+3/x^2+x-2 ÷ 4x^2-12x+9/x^2-4 I got 3x^3+2x^2-12x+8 / (x-1)(2x-3) (x-4) Is this right please show your steps so I can understand! Thanks I greatly appreciate your help! Iwas just wondering wen my

More Similar Questions