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nasty-shy [4]
3 years ago
13

Find values of x and y for which ABCD must be a parallelogram. The diagram is not to

Mathematics
2 answers:
Nutka1998 [239]3 years ago
8 0
3x-14=x+2
Add 14 to both sides.
3x=x+16
Subtract x from both sides.
2x=16
Divide by 2.
x=8

4y-7=y+11
Add 7 to both sides.
4y=y+18
Subtract y from both sides.
3y=18
Divide by 3.
y=6

The answer is d.

Zina [86]3 years ago
7 0
The answer is d. 

3x-14 = x-2     solve for x

4y-7 = y+11    solve for y
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Let x, y and z denote the weighs of car X, car Y and car Z, respectively.

We know that car X weighs 136 more than car Z, this can be express by the equation:

x=z+136

We also know that Y weighs 117 pounds more than car Z, this can be express as:

y=z+117

Finally, we know that the total weight of all the cars is 9439, then we have:

x+y+z=9439

Hence, we have the system of the equations:

\begin{gathered} x=z+136 \\ y=z+117 \\ z+y+z=9439 \end{gathered}

To solve the system we can plug the values of x and y, given in the first two equations, in the last equation; then we have:

\begin{gathered} z+136+z+117+z=9439 \\ 3z=9439-136-117 \\ 3z=9186 \\ z=\frac{9186}{3} \\ z=3062 \end{gathered}

Now that we have the value of z we plug it in the first two equations to find x and y:

\begin{gathered} x=3062+136=3198 \\ y=3062+117=3179 \end{gathered}

Therefore, car X weighs 3198 pound, car Y weighs 3179 pounds and car Z weighs 3062 pounds.

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