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2 years ago

11

For the following pairs of equations explain why the equations are NOT equivalent.

1 answer:

Serga [27]2 years ago

7 0

If you were to make both x's equivalent, you'd have to multiply the first equation by 2 to get rid of the coefficient, 1/2. However, you'd have to multiply 2 onto -8 as well, therefore the equation would turn into x - 16 = 18. That is not equivalent to x - 8 = 18.

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Graph the system of equations.<br><br> {2x - y =4<br> {x - y = -2

Mnenie [13.5K]

Answer:

See attachment

Step-by-step explanation:

The given system is

2x-y=4

x-y=-2

We write the system in slope-intercept form:

y=2x-4

y=x+2

The first equation has slope of 2 and y-intercept at (0,-4)

When y=0, x=2.

We plot (0,-4) and (2,0) to graph this first equation.

The second equation has a slope of 1 and a y-intercept at (0,2)

Also when y=0, x=-2.

We plot (0,2) and (-2,0) and draw a straight line through them.

The graph of the system is shown in the attachment.

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3 years ago

Find the measure of the missing angles! Pls help thank you

Inga [223]

Answer:

a=61°

Step-by-step explanation:

a= 180° - 90° - 29°

a = 61°

3 0

2 years ago

What is the multiplicative inverse of -1/3

motikmotik

-1/3 . -3/1 = 1
......................

4 0

3 years ago

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1 year ago

Read 2 more answers

Fill in the missing term in the equation

Y_Kistochka [10]

Answer:

<u>x</u>

Step-by-step explanation:

Let's simplify the LHS so it can be equated to the RHS.

$\frac{x}{x^{2} -4x+4} - \frac{x}{x^{2} -3x+2}$

$\frac{x}{(x-2)^{2} } - \frac{x}{(x-1)(x-2)}$

$\frac{x(x-1)-x(x-2)}{(x-2)^{2}(x-1)}$

$\frac{x^{2} -x-x^{2}+2x }{(x-2)^{2}(x-1)}$

$\frac{x}{(x-2)^{2}(x-1)}$

Hence, the missing term in the numerator of the answer is <u>x</u>

6 0

1 year ago