Answer:
this may help you.look it once.
Since PQ bisects CX at point J, JX=CJ
4y=y+9
-y -y
3y=9
divide by 3
y=3
Answer: D
Step-by-step explanation:
Consider the first equation. Subtract 3x from both sides.
y−3x=−2
Consider the second equation. Subtract x from both sides.
y−2−x=0
Add 2 to both sides. Anything plus zero gives itself.
y−x=2
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
y−3x=−2,y−x=2
Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.
y−3x=−2
Add 3x to both sides of the equation.
y=3x−2
Substitute 3x−2 for y in the other equation, y−x=2.
3x−2−x=2
Add 3x to −x.
2x−2=2
Add 2 to both sides of the equation.
2x=4
Divide both sides by 2.
x=2
Substitute 2 for x in y=3x−2. Because the resulting equation contains only one variable, you can solve for y directly.
y=3×2−2
Multiply 3 times 2.
y=6−2
Add −2 to 6.
y=4
The system is now solved.
y=4,x=2
for 
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Yes. This is true.