ABCD is a parallelogram. if side AB=6x-10, side BC=4x-5, and side CD =3x+5, find the value of x

Mathematics

1 answer:

serg [7]3 years ago

8 0

Answer:

x=5

Step-by-step explanation:

If ABCD is a parallelogram, then AB = CD

AB=CD

6x-10 = 3x+5

Subtract 3x from each side

6x-3x -10 = 3x-3x+5

3x-10 = 5

Add 10 to each side

3x-10+10 = 5+10

3x = 15

Divide by 3 on each side

3x/3 =15/3

x=5

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It x, y, and z are positive intergers and 3x = 4y = 7z, then the least possible value of x + y + z is???

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$3x = 4y = 7z \\ \Rightarrow \begin{cases}3x - 4y = 0 \\4y - 7z = 0 \\ 7z - 3x = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 3( \frac{4}{3}y) = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 4y= 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\ y \\ z = \frac{4}{7}y \end{cases} \\ x,y,z\: are\: intergers\Rightarrow y=LCM(3,7)=21\Rightarrow x = 28,z = 12 \\ \Rightarrow x + y + z = 61$