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Rectangle upper A upper B upper C upper D is shown with point upper O in the center. Diagonal segments upper A upper C and upper B upper D intersect at point upper O, in the center of the rectangle. Some line segments have measurements.

Given the rectangle ABCD as shown, find the value for x if DO=4x+8 and CO=15x−3 .

(1 point)
x=

1 answer

In rectangle \( ABCD \), the diagonals \( AC \) and \( BD \) intersect at point \( O \), which is the midpoint of both diagonals. Therefore, the segments \( DO \) and \( CO \) are equal:

\[ DO = CO \]

Given: \[ DO = 4x + 8 \] \[ CO = 15x - 3 \]

Setting these two expressions equal to each other, we have:

\[ 4x + 8 = 15x - 3 \]

To solve for \( x \), first, we will rearrange the equation:

  1. Subtract \( 4x \) from both sides:

\[ 8 = 15x - 4x - 3 \]

  1. Simplifying gives:

\[ 8 = 11x - 3 \]

  1. Now, add 3 to both sides:

\[ 8 + 3 = 11x \] \[ 11 = 11x \]

  1. Finally, divide both sides by 11:

\[ x = 1 \]

So, the value for \( x \) is:

\[ \boxed{1} \]