Question

2+5x = 11 + 8x+3
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4 28 7

Answer 1

x= 3

Answer 2

would you be able to kind of work out the problem for me? I would appreciate lots

Answer 3

u wanted to find x right, then sure

Answer 4

samsterz, do not give direct answers. The lowest common denominator is 28 so multiply everything by that. This would give you 7(2+5x) = 11 + 4(8x+3). From there you can distribute and get 14 + 35x = 11 +32x + 12 which would become 14+35x = 23 + 32x. Get variable on one side by manipulating with subtraction. You will get 14 + 3x = 23. Now get numbers without variables to other side. You will get 3x = 9. Get x alone, x=3.

Answer 5

ash vasan i said id give an explanation

Answer 6

ok back

Answer 7

Multiply each term by a factor of 1 that will equate all the denominators. In this case, all terms need a denominator of 28. The (8x+3)7 expression needs to be multiplied by (4)(4) to make the denominator 28.


1/4(2+5x)=1128+17(8x+3)⋅44

Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 28.

1/4(2+5x)=1128+(8x+3)(4)28

Multiply 4 by each term in 8x+3 to get 32x+12.


Multiply each term in the first polynomial by each term in the second polynomial.

1/4(2+5x)=1128+8x⋅4+3⋅428


Multiply 8x by 4 to get 32x.

1/4(2+5x)=1128+128(32x+3⋅4)


Multiply 3 by 4 to get 12.

1/4(2+5x)=1128+128(32x+12)

1/4(2+5x)=1128+128(32x+12)

The numerators of expressions that have equal denominators can be combined. In this case, 1128 and (32x+12)28 have the same denominator of 28, so the numerators can be combined.

1/4(2+5x)=11+(32x+12)28

Simplify the numerator of the expression.

1/4(2+5x)=128(11+32x+12)

Add 12 to 11 to get 23.

1/4(2+5x)=128(23+32x)

Reorder the polynomial 23+32x alphabetically from left to right, starting with the highest order term.

1/4(2+5x)=128(32x+23)

Reorder the polynomial 2+5x alphabetically from left to right, starting with the highest order term.

1/4(5x+2)=128(32x+23)

Since there is one rational expression on each side of the equation, this can be solved as a ratio. For example, AB=CD is equivalent to A⋅D=B⋅C.

(5x+2)⋅28=(32x+23)⋅4

Multiply 4 by each term inside the parentheses.

More Detail
Multiply (32x+23) by 4 to get 4(32x+23).

(5x+2)⋅28=4(32x+23)


Multiply 4 by each term inside the parentheses (32x+23).

(5x+2)⋅28=4(32x)+4(23)


Multiply 4 by the 32x inside the parentheses.

(5x+2)⋅28=4⋅32x+4(23)


Multiply 4 by 32x to get 128x.

(5x+2)⋅28=128x+4(23)


Multiply 4 by the 23 inside the parentheses.

(5x+2)⋅28=128x+4⋅23


Multiply 4 by 23 to get 92.

(5x+2)⋅28=128x+92

(5x+2)⋅28=128x+92

Multiply 28 by each term inside the parentheses.


Multiply (5x+2) by 28 to get 28(5x+2).

28(5x+2)=128x+92


Multiply 28 by each term inside the parentheses (5x+2).

28(5x)+28(2)=128x+92


Multiply 28 by the 5x inside the parentheses.

28⋅5x+28(2)=128x+92


Multiply 28 by 5x to get 140x.

140x+28(2)=128x+92


Multiply 28 by the 2 inside the parentheses.

140x+28⋅2=128x+92


Multiply 28 by 2 to get 56.

140x+56=128x+92

140x+56=128x+92

Since 128x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 128x from both sides.

140x+56−128x=92

Since 140x and −128x are like terms, add −128x to 140x to get 12x.

12x+56=92

Since 56 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 56 from both sides.

12x=−56+92

Add 92 to −56 to get 36.

12x=36

Divide each term in the equation by 12.



Simplify the left-hand side of the equation by canceling the common factors.

Cancel the common factor of 12 in 12x/12.




Reduce the expression 12x/12 by removing a factor of 12 from the numerator and denominator.

x=36/12

Simplify the equation.


Cancel the common factor of 12 in 36/1212.


Reduce the expression 3612 by removing a factor of 12 from the numerator and denominator.

Answer 8

and u get x=3

Answer 9

a lot of thanks to both