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3 years ago

What is the value of x in the equation 2(6x+4)-6+2x=3(4x+3)+1? 1 3 4 6

Mathematics

2 answers:

sp2606 [1]3 years ago

6 0

Here is the process I did:

2(6x+4)-6+2x=3(4x+3)+1

Step 1: Distribute the 2 to the numbers in the parentheses (6x +4) on the left side of the equation

6x(2)+4(2)-6+2x=3(4x+3)+1

12x+8-6+2x=3(4x+3)+1

Step 2: On the left side of the equation combine like terms

x's go with x's (12x and 2x):

(12x + 2x)+8-6=3(4x+3)+1

14x+8-6=3(4x+3)+1

normal numbers go with normal numbers ( 8 and -6):

14x+(8+(-6))=3(4x+3)+1

14x+2=3(4x+3)+1

Step 3: Distribute the 3 to the numbers in the parentheses (4x+3) on the right side of the equation

14x+2=4x(3) +3(3) + 1

14x+2=12x + 9 + 1

Step 4: On the right side of the equation combine like terms

normal numbers go with normal numbers ( 9 and 1):

14x+2=12x + (9 + 1)

14x+2=12x + 10

Step 5: Bring 2 from the left side to the right by subtracting it to both sides

14x+(2-2)=12x + (10-2)

14x = 12x + 8

Step 6: Bring 12x from the right side to the left by subtracting it to both sides

(14x - 12x) = (12x - 12x) + 8

2x = 8

Step 7: Isolate x by dividing 2 to both sides

$\frac{2x}{2} =\frac{8}{2}$

x = 4

Hope this helped!

german3 years ago

3 0

Answer:

Step-by-step explanation:

2(6x+4)-6+2x=3(4x+3)+1

Step 1: Distribute the 2 to the numbers in the parentheses (6x +4) on the left side of the equation

6x(2)+4(2)-6+2x=3(4x+3)+1

12x+8-6+2x=3(4x+3)+1

Step 2: On the left side of the equation combine like terms

x's go with x's (12x and 2x):

(12x + 2x)+8-6=3(4x+3)+1

14x+8-6=3(4x+3)+1

Read more on Brainly.com - brainly.com/question/12681318#readmore

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One positive number is 14 less than another positive number. If the reciprocal of the smaller number is added to five times the

Kipish [7]

Let's call the two numbers $s$ and $l$ .

Given these variables, we can say: $s = l - 14$ , based on the first sentence in the problem.

Also, remember that the reciprocal of a number is simply 1 divided by the number. Thus, we can say that:

$\dfrac{1}{s} + 5\Big( \dfrac{1}{l} \Big) = \dfrac{1}{4}$

To solve, we can simply substitute $l -14$ in for $s$ in the second equation and solve.

$\dfrac{1}{l - 14} + \dfrac{5}{l} = \dfrac{1}{4}$

Set up

$\dfrac{l}{l(l - 14)} + \dfrac{5(l - 14)}{l(l - 14)} = \dfrac{1}{4}$

Get terms on the left side to a common denominator for easier addition

$\dfrac{l + 5l - 70}{l(l - 14)} = \dfrac{1}{4}$

Simplify

$4(6l - 70) = l(l - 14)$

Cross multiplication ( $\frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc$ )

$24l - 280 = l^2 - 14l$

Simplify

$l^2 - 38l + 280 = 0$

Subtract $24l - 280$ from both sides of the equation

$(l - 10)(l - 28) = 0$

Factor left side of the equation

$l = 10, 28$

Zero Product Property

Now, notice that we have found two solutions, but the problem is only asking for one. This <em>likely </em>means that one of our solutions is extraneous. Let's take a look. Remember that the smaller positive number is equal to 14 less than the larger number. However,

$s = 10 - 14 = -4$ ,