In AJKL and APQR, if JK. PO, LJ - Z. P. and ZK2Q, then AJKL must be
congruent to APQR.

True → A

Step-by-step explanation:

Let us revise the cases of congruence  

SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd ΔASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd ΔAAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd ΔHL ⇒ hypotenuse and leg of the 1st right Δ ≅ hypotenuse and leg of the 2nd right Δ.  

In the two Δs JKL and PQR

∵ JK ≅ PQ

∵ ∠J ≅ ∠P

∵ ∠K ≅ ∠Q

∵ Side JK is joining the angles J and K

∵ Side PQ is joining the angles P and Q

→ By using case 3 above ⇒ ASA postulate of congruency

∴ Δ JKL must congruent to Δ PQR

∴ The answer is true

2x + 3y = 114x + 2y = - 2 ( original equation x -1)4x + 6y = 22 ( original equation x 2)- 4x + 2y = - 2= 4y = 24÷4y = 62x + 18 = 11-182x = - 7÷2x = - 3.5