Solve for x:
(x + 15)/5 = (7 - x)/3 + 1/2
Put each term in (7 - x)/3 + 1/2 over the common denominator 6: (7 - x)/3 + 1/2 = 3/6 + (2 (7 - x))/6:
(x + 15)/5 = 3/6 + (2 (7 - x))/6
3/6 + (2 (7 - x))/6 = (2 (7 - x) + 3)/6:
(x + 15)/5 = (2 (7 - x) + 3)/6
2 (7 - x) = 14 - 2 x:
(x + 15)/5 = (14 - 2 x + 3)/6
Add like terms. 3 + 14 = 17:
(x + 15)/5 = (17 - 2 x)/6
Multiply both sides by 30:
6 (x + 15) = (30 (17 - 2 x))/6
30/5 = (5×6)/5 = 6:
6 (x + 15) = (30 (17 - 2 x))/6
30/6 = (6×5)/6 = 5:
6 (x + 15) = 5 (17 - 2 x)
Expand out terms of the left hand side:
6 x + 90 = 5 (17 - 2 x)
Expand out terms of the right hand side:
6 x + 90 = 85 - 10 x
Add 10 x to both sides:
10 x + 6 x + 90 = (10 x - 10 x) + 85
16 x + 90 = 85
Subtract 90 from both sides:
16 x + (90 - 90) = 85 - 90
16 x = -5
Divide both sides of 16 x = -5 by 16:
x = (-5)/16
+590 First, let's distribute the left side of the equation
1/5x+3=1/2-1/3(x-7)
Then let's distribute 1/3 to (x-7) on the right side
1/5x+3=1/2-1/3x-7/3
There are too many fractions...so multiply 30 to both sides. 30 is the LCM of 2, 3, and 5.
Then we get 6x+90=15-10x+70. I admit there are other ways to do it...
Add 10x to both sides to get 16x +90=15+70
Simplify to get 16x+90=85
Subtract 90 to both sides and get 16x=-5
Divide both sides by 16 and get x=-5/16
x=-5/16.
(Oh, I see someone wrote the answer while I was typing this. I'll post this anyway...
)
+26393 solve for x
1/5(x+15)=1/2-1/3(x-7)
\(\begin{array}{|rcll|} \hline \frac15(x+15) &=& \frac12- \frac13(x-7) \quad & | \quad \times 5 \\ \frac55(x+15) &=& \frac52- \frac53(x-7) \\ x+15 &=& \frac52- \frac53(x-7) \quad & | \quad \times 3 \\ 3(x+15) &=& \frac{5\cdot 3}{2}- \frac{5\cdot 3}3(x-7) \\ 3(x+15) &=& \frac{15}{2}- 5(x-7) \quad & | \quad \times 2 \\ 2\cdot 3(x+15) &=& \frac{15\cdot 2}{2}- 2\cdot 5(x-7) \\ 6(x+15) &=& 15- 10(x-7) \\ 6x+6\cdot 15 &=& 15- 10x -10(-7) \\ 6x+90 &=& 15- 10x +70 \quad & | \quad + 10x \\ 10x+6x+90 &=& 15 +70 \\ 16x+90 &=& 85 \quad & | \quad -90\\ 16x &=& 85 -90 \\ 16x &=& -5 \quad & | \quad : 16 \\ \mathbf{x} & \mathbf{=} & \mathbf{-\frac{5}{16}} \\ \hline \end{array}\)
