PQ = 3x + 14 and QR = 7x 10; Find x.

Answer 1

PQ = 3x+14

QR = 7x +10

if Q is the midpoint, 3x+14 =7x+10

4x =4

x =1

PQ =17, QR =17

PR =2(17) =34

or maybe you meant

QR =7x-10

then 3x+14 =7x-10

7x-3x = 14+10

4x =24

x =24/4 =6

PQ =3(6)+14 = 32

QR =7(6)-10 = 32

PR = 64

or maybe you meant

QR =7x^10

then 3x+14 =7x^10

which has 10 solutions for x, 8 imaginary, 2 real

x= about -1.04494 and 1.09459

3x-14 = about 10.865 or 17.284

PR =2(10.865) = 21.73 or 2(17.284) = 34.57

graph the line 3x+14 and the 10 degree 7^10

and find points of intersection. Use a graphing calulator

or maybe PQ and QR are two sides of a triangle

possibly an equilateral triangle

Then PR=PQ=QR

or maybe it's not an equilateral triangle

or maybe PQ and QR are in some other ratio of lengths

there's an infinite number of possibilities

but the 1st two intepretations of the problem came out so evenly, to similar integers for x, it's likely you intended one of those two problems with x=1, or x=6

another perspective could be to graph

y=3x+14

and

y=7x +10

to find their point of intersection (1,17)

then PR = 2(17) = 34

or y=3x+14 and y=7x+10 intersect at (6,32)

PR = 2(32) = 64