Find: a) QR
b) 𝑚∠𝑃𝑄𝑇

Solution :-
Hint :- Given, Total income of the store for 70 soccer balls is $2,400.
There are two types of soccer balls and the cost of each type is different .
Frame equation considering no.of limited edition soccer balls sold be x
And no.of Pro NSL soccer ball sold be y and solve them to find  x and y.
Ans :- The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.
Explanation :-
Let no.of limited edition soccer balls sold be x ,no.of Pro NSL soccer ball sold be y.
Step 1:- Frame equations
Total no.of ball is 70
I.e x + y = 70             —Eq1
Total cost of balls is $2,400
Cost of x  limited edition soccer balls is 65x (as per ball cost is given in diagram)
And Cost of y Pro NSL soccer ball is 15x(as per ball cost is given in diagram)
I.e 65x + 15y = 2,400            —Eq2
Step 2:- Eliminate y to find x
Do Eq2 -15(Eq1) to eliminate y
65x + 15y - 15(x+y) = 2400 - 15(70)
65x - 15x = 1350
50x = 1350 ⇒ x = 27
Step 3:- substitute value of x to find y
x + y = 70 ⇒ 27 + y = 70
⇒ y = 70 - 27
∴y = 43
∴The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.

• Step by step explanation: 
• Given:

𝑚∠POQ = (x - 4) °  

𝑚∠QOR = (5x - 20) °.

• Step 1:
• From the figure it is clear that,

∠POR is right angle hence ∠POR = 90^o.
and 

∠POR = ∠POQ + ∠QOR 

• Step 2:
• Put values of ∠COD and ∠COP 

∠POR = ∠POQ + ∠QOR  

90 = (x - 4) + (5x - 20)

90 = 5x + x - (20 + 4)

90 = 6x - 24

6x = 90 - 24

6x = 66

x = 

x =  11

• Final Answer:

Hence, the x is 11.

• Step by step explanation: 

• Step 1:

Find pair of congruent angles
From the figure it is clear that

∠BAC = ∠QPR

∠BCA = ∠QRP

∠ABC = ∠PQR
This are pair of congruent angles.

• Step 2:

Find pair of congruent segments
From the figure it is clear that

BA = QP                          
This is pair of congruent segments.

HINT – If sum of two angles is 90°, we say they are complementary angles
SOL – It is given that ∠𝐴 & ∠𝐵 are complementary
∠A + ∠B = 90°               ---- (1)
Also, 𝐵 & ∠𝐶 are complementary
 ∠B + ∠C = 90°              ---- (2)
From (1) and (2)
We get, ∠A + ∠B = ∠B + ∠C
 ∠A = ∠C
 ∠𝐴 ≅  ∠𝐶
Hence Proved.

SOL – It is given that ray OR bisects ∠𝑃𝑂𝑆 i.e. OR is an angle bisector.
We know that an angle bisector divides an angle into two congruent angles.
∠ POR ≅ ∠ ROS
 𝑚∠1 = 𝑚∠2
Hence Proved.

Ans :- x = 1 ; y = -5
Explanation :-
⇒  y = -2x - 3 — eq 1
⇒  y = -x - 4—- eq 2
Step 1 :- find x by substituting y = 4x + 2 in eq 2.
-2x - 3 =   -x – 4  ⇒  2x + 3x + 4
⇒  2x – x = 4 - 3 ⇒ x  = 4 - 3
⇒  x  = 1
Step 2 :- substitute value of x and find y
⇒  y =  - x – 4  ⇒  y = -1 - 4
∴  y =  - 5
∴   x = 1 and y = - 5  is the solution of the given pair of equations.

SOLUTION:
HINT: Perform any arithmetic operation and then find.
Complete step by step solution:
Let x - 2y = 1…(i)
and 2x + 3y=-12….(ii)
On multiplying (i) with 2, we get 2(x - 2y = 1)
⇒ 2x - 4y = 2…(iii)
Now, we have the coefficients of x in (ii) and (iii) to be the same.
On subtracting (ii) from (iii),
we get LHS to be 2x - 4y - (2x + 3y) = - 4y - 3y = - 7y
and RHS to be 2 - (- 12) = 14
On equating LHS and RHS, we have - 7y = 14
⇒ y = - 2
On substituting the value of y in (i), we get x - 2 × - 2 =1
⇒ x + 4 = 1
⇒ x = 1-4
⇒ x = - 3
Hence we get x = - 3 and y = - 2
Note: We can also solve these system of equations by making the coefficients of y
to be the same in both the equations

HINT – Open the brackets
SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
Opening the brackets
We get, x – 2 + 3x + 6 = 3x + 10
4x + 4 = 3x + 10      ( Adding similar terms )
 4x – 3x = 10 – 4
 x = 6.

Solution :-
Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of  x in the equations  .If we get a true statement we say they have infinite solutions .If we get the false statement we say they have no solution.
Ans :- infinite no.of solutions .
Explanation :-
y = 2x + 2— eq 1
6x - 3y = -6—- eq 2
Step 1 :- find x by substituting y = 2x + 2 in eq 2.
6x - 3 (2x + 2) = -6 ⇒ 6 x -6x - 6 = -6
-6 = -6
Here we get -6 = -6 which is always true i.e always having a root .
They coincide with each other and have infinite no.of solutions
They have infinite no.of solutions for the given system of equations

SOL – It is given that 𝑚∠𝑃 = 30° and  𝑚∠𝑄 = 30°

𝑚∠𝑃 = 𝑚∠𝑄           ---- (1)
Further, it is given that 𝑚∠𝑄 = 𝑚∠𝑅    ---- (1)
Using transitive property which states that if A = B and B = C then A = C
We get from (1) and (2),

𝑚∠P = 𝑚∠R = 30°
Hence Proved

HINT – Use the information given.
SOL – It is given that 𝑚∠𝑃𝑀𝐶 = 150°    ---- (1)
𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180°
150° + 𝑚∠𝑃𝑀𝐷 = 180°             ( - From (1) )
𝑚∠𝑃𝑀𝐷 = 180° - 150°
= 30°
Hence Proved.

Complete step by step solution:
Let 3x + 2y = 8…(i)
and x + 4y = - 4….(ii)
On multiplying (ii) with 3, we get 3(x + 4y=-4)
⇒3x + 12y = - 12…(iii)
Now, we have the coefficients of  in (i) and (iii) to be the same.
On subtracting (i) from (iii),
we get LHS to be 3x + 12y - (3x + 2y) = 12y - 2y = 10y
and RHS to be - 12 - 8 = - 20
On equating LHS and RHS, we have 10y = - 20
⇒y = - 2
On substituting the value of y in (i), we get 3x + 2× - 2 = 8
⇒ 3x - 4 = 8
⇒ 3x = 8 + 4
⇒ 3x = 12
⇒x = 4
Hence we get x = 4 and y = - 2
Note: We can also solve these system of equations by making the coefficients of y
to be the same in both the equations

HINT – Use Segment Addition Postulate
SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR
PR = PQ + QR
 PR = 12 + 13
 PR = 25 in.
Hence Proved.

Solution :-
Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of  x in the equations.
Ans :- x = -4; y = 5
Explanation :-
- 3x - y = 7 ⇒ - 3x - 7 = y                     — eq 1
x + 2y = 6                                     —- eq 2
x + 2 ( - 3x - 7) = 6 ⇒ x - 6x - 14 = 6
⇒ - 5x = 14 + 6 ⇒ - 5x = 20
⇒ x = -4
Step 2 :- substitute value of x and find y
⇒ - 3x - 7 = y ⇒ y = - 3 (- 4) - 7
⇒ y = 12 - 7
∴ y = 5
x = -4 and y = 5 is the solution of the given pair of equations.

HINT – Use Segment Addition Postulate
SOL – If Q is mid – point  Þ PQ = QR      ---- (1)
Using Segment Addition Postulate,
We get,  PR = PQ + QR
PR = PQ + PQ
PR = 2 PQ
Hence Proved.