Question

if r is the midpoint of qs, rs=2x-4, st=4x-1, and rt=8x-43, find qs

Answer 1

Answer:

$\large\boxed{QS=22\dfrac{2}{3}}$

Step-by-step explanation:

Look at the picture.

We must start from domain:

$1)\\2x-4>0$    add 4 to both sides

$2x>4$     divide both sides by 2

$x>2$

and

$2)\\4x-1>0$          add 1 to both sides

$4x>1$          divide both sides by 4

$x>\dfrac{1}{4}$

and

$3)\\8x-43>0$            add 43 to both sides

$8x>43$           divide both sides by 8

$x>\dfrac{43}{8}\\\\x>5\dfrac{3}{8}$

From 1, 2 and 3 we have:

$D:x>5\dfrac{3}{8}$

$\bold{(1)}\\\\(8x-43)+(4x-1)=2x-4\\8x-43+4x-1=2x-4\qquad\text{combine like terms}\\(8x+4x)+(-43-1)=2x-4\\12x-44=2x-4\qquad\text{add 44 to both sides}\\12x=2x+40\qquad\text{subtract}\ 2x\ \text{from both sides}\\10x=40\qquad\text{divide both sides by 10}\\x=4\notin D$

$\bold{(2)}\\\\2x-4=(4x-1)-(8x-43)\\2x-4=4x-1-8x-(-43)\\2x-4=(4x-8x)+(-1+43)\qquad\text{combine like terms}\\2x-4=-4x+42\qquad\text{add 4 to both sides}\\2x=-4x+46\qquad\text{add}\ 4x\ \text{to both sides}\\6x=46\qquad\text{divide both sides by 6}\\x=\dfrac{46}{6}\\\\x=\dfrac{23}{3}\\\\x=7\dfrac{2}{3}\in D$

R is midpoint of QS. Therefore QR = RS ⇒ QS = 2(RS).

RS = 2x - 4

Put the value of x and calculate the length of RS:

$RS=2\left(\dfrac{23}{3}\right)-4=\dfrac{46}{3}-\dfrac{12}{3}=\dfrac{34}{3}$

Therefore

$QS=2\left(\dfrac{34}{3}\right)=\dfrac{68}{3}=22\dfrac{2}{3}$