All categories

3 years ago

Negative 4 (2 x + 3) = 2 x + 6 minus (8 x + 2)

Mathematics

1 answer:

jekas [21]3 years ago

4 0

Answer:

x=−8

Step-by-step explanation:

−4(2x+3)=2x+6−(8x+2)

−4(2x+3)=2x+6+−1(8x+2) (Distribute the Negative Sign)

−4(2x+3)=2x+6+−1(8x)+(−1)(2)

−4(2x+3)=2x+6+−8x+−2

(−4)(2x)+(−4)(3)=2x+6+−8x+−2 (Distribute)

−8x+−12=2x+6+−8x+−2

−8x−12=(2x+−8x)+(6+−2) (Combine Like Terms)

−8x−12=−6x+4

−8x−12=−6x+4 Add 6x to both sides.

−8x−12+6x=−6x+4+6x

−2x−12=4 Add 12 to both sides.

−2x−12+12=4+12

−2x=16 Divide both sides by -2.

−2x

−2

x=−8

You might be interested in

Helppp!! Malcolm is training for a marathon. He runs 3 miles in

Arlecino [84]

Answer:

Malcolm would logically be running at 4.75 and since he is running at 3, it would be the lowest possible therefore its B

Step-by-step explanation:

7 0

3 years ago

Jill was solving a system using the elimination method.After she added the equations together,she got 5=7.What does this mean fo

harkovskaia [24]

Answer:

It means the system does not have a solution

Step-by-step explanation:

5 0

3 years ago

The options are greater than less than, at most and at least

Dmitrij [34]

Answer:

it is unable to answered i cant see the picture

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Graph the image of this figure after a dilation with a scale factor of 13 centered at the point (4, −2) .

Sonja [21]

Answer:

From the given triangle figure;

Labelled the triangle as A , B and C.

The coordinates of this triangle ABC are;

A = (1, 10) ,

B = (-2, 4)

C = (7, 4).

Given : Scale factor(k) = $\frac{1}{3}$ and centered at point (4, -2).

The rule of dilation with $k= \frac{1}{3}$ and center at point (4,-2) is:

$(x, y) \rightarrow (\frac{1}{3}(x-4)+4 , \frac{1}{3}(y+2)-2)$

$(x, y) \rightarrow (\frac{1}{3}x-\frac{4}{3}+4 , \frac{1}{3}y+\frac{2}{3}-2)$

$(x, y) \rightarrow (\frac{1}{3}x+\frac{8}{3} , \frac{1}{3}y-\frac{4}{3})$

then, the dilation of the given figure are;

$A(1, 10) \rightarrow (\frac{1}{3}\cdot 1+\frac{8}{3} , \frac{1}{3} \cdot 10-\frac{4}{3})$ = $(\frac{1}{3}+\frac{8}{3} , \frac{10}{3}-\frac{4}{3})$ = $(\frac{9}{3} , \frac{6}{3})$ = A'(3 , 2)

$B(-2, 4) \rightarrow (\frac{1}{3}\cdot -2+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})$ = $(-\frac{2}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})$ = $(\frac{6}{3} , \frac{0}{3})$ =B'(2 , 0)

$C(7, 4) \rightarrow (\frac{1}{3}\cdot 7+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3})$ = $(\frac{7}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3})$ = $(\frac{15}{3} , \frac{0}{3})$ = C'(5 , 0)