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2 years ago

2(x)+2(2)=2(x/2)-2(x) Solve this please and show the steps!!!!!!

Mathematics

2 answers:

FinnZ [79.3K]2 years ago

7 0

Answer:-4/3

Step-by-step explanation:

2(x)+2(2)=2(x/2)-2(x)

Remove brackets by multiplying through

2x + 4=x - 2x

Collect like terms

2x+2x-x=-4

3x=-4

Divide both sides by 3

3x/3=-4/3

x=-4/3

Olin [163]2 years ago

5 0

Answer: $x=-\frac{4}{3}$

Step-by-step explanation:

$2(x)+2(2)=2(\frac{x}{2})-2(x)$

Remove all those parentheses by multiplying. Note: 2 and 2 cancel out.

$2x+4=x-2x$

Subtract 2x

$4=x-2x-2x$

Combine like terms;

$4=-3x$

Divide by -3

$\frac{4}{-3}=x$

Rewriting it;

$x=-\frac{4}{3}$

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<h2>Solving Equations</h2>

To solve linear equations, we must perform inverse operations on both sides of the equal sign to <em>cancel values out</em>.

If something is being added to x, subtract it from both sides.
If something is being subtracted from x, add it on both sides.
Same with multiplication and division. If x is being divided, multiply. If x is being multiplied, divide.

We perform inverse operations to<em> combine like terms</em>. This means to get x to one side and everything else on the other.

<h2>Solving the Questions</h2><h3>Question 1</h3>

$5x+7=15$

Because 7 is being added to x, subtract it from both sides:

$5x+7-7=15-7\\5x=8$

Because x is being multiplied by 5, divide both sides by 5:

$\dfrac{5x}{5}=\dfrac{8}{5}\\\\x=\dfrac{8}{5}$

Therefore. $x=\dfrac{8}{5}$ .

<h3>Question 2</h3>

$7x+4=5x-18$

Here, we can group all the x values on the left side of the equation. Subtract 5x from both sides:

$7x+4-5x=5x-18-5x\\2x+4=-18$

To isolate x, subtract 4 from both sides:

$2x+4-4=-18-4\\2x=-22$

Divide both sides by 2:

$\dfrac{2x}{2}=\dfrac{-22}{2}\\\\x=-11$

Therefore, $x=-11$ .

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2 years ago

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3 years ago

find the component equation of the plane which is normal to the vector -2i+5j+k and which contains the point (-10;7;5).

maksim [4K]

Given:

A plane is normal to the vector = -2i+5j+k

It contains the point (-10,7,5).

To find:

The component equation of the plane.

Solution:

The equation of plane is

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

Where, $(x_0,y_0,z_0)$ is the point on the plane and $\left< a,b,c\right>$ is normal vector.

Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is

$-2(x-(-10))+5(y-7)+1(z-5)=0$

$-2(x+10)+5y-35+z-5=0$

$-2x-20+5y-35+z-5=0$

$-2x+5y+z-60=0$

$-2x+5y+z=60$

Therefore, the equation of the plane is $-2x+5y+z=60$ .

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2 years ago