Use Substitution to solve each system of equations :
6X - 3Y = -6
Y = 2X + 2

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.

• Step by step explanation: 
• Step 1:
• Find ∠POQ:

• Align the protractor with the ray OP on 0^o as shown above. 
• Start reading the inner scale from the 0°.
• Step 2:
• From the figure we can see that ray OP is aligned on mark 0^o. And ray OQ is aligned on mark 30^o.

Hence, the measure of ∠POQ = (30^o - 0^o).
∠POQ = 30^o.

• Step 3:
• Find ∠SOT:
  • Align the protractor with the ray OT on 0^o as shown above. 
  • Start reading the outer scale from the 0°.
  • Step 4:
  • From the figure we can see that ray OT is aligned on mark 0^o. And the ray OS is aligned on mark 20^o.

Hence, the measure of ∠SOT = 20^o.

∠SOT = 20^o.

• Step 5:
• Compare ∠POQ and ∠SOT

Hence, ∠POQ is greater than ∠SOT by 10^o.

• Final Answer:

Hence, ∠POQ is greater than ∠SOT by 10^o.

HINT – Try to recall definition of postulates/properties given in options.
SOL – Option (b)
Segment addition postulate – It states that if a line segment has two endpoints, A and C, a third point B lies on the line segment AC if and only if this equation AB + BC = AC
Since in figure given above, line segment has two end points namely P and R and third point Q lies on the line segment PR Þ PR = PQ + QR

SOLUTION:
HINT: Perform any arithmetic operation and then find.
Complete step by step solution:
Let x - 2y = - 2…(i)
and 3x + 2y = 30….(ii)
On adding (i) and (ii),
we get LHS to be x - 2y + 3x + 2y = 4x
and RHS to be -2+ 30 = 28
On equating LHS and RHS, we get 4x = 28
⇒ x = 7
On substituting the value of x in (i), we get 7 - 2y = - 2
⇒ - 2y = - 2 - 7
⇒ - 2y = - 9
⇒ y =
Hence we get x = 7 and 
Note: We can also solve these system of equations by making the coefficients of x
to be the same in both the equations.

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 = 2 ; y =10
Explanation :-
⇒  y =  4x + 2 — eq 1
⇒  y  = x + 8—- eq 2
Step 1 :- find x by substituting  y = 4x + 2 in eq 2.
4x + 2 = x + 8 ⇒  4x – x = 8 - 2
⇒  3x = 8  - 2  ⇒ 3x = 6
⇒ x =2
Step 2 :- substitute value of x and find y
⇒  y =  x + 8 ⇒ y = 2 + 8
∴ y= 10
∴  x = 2 and y = 10  is the solution of the given pair of equations.

SOLUTION:
HINT: Use the property of parallel lines angle rules.
Complete step by step solution:
Here we have 2 parallel lines m and land a transversal intersecting these parallel
lines.
Here,   forms co-interior angles and they add up to 
(co-interior angles)




Hence the value of 

HINT – Use Euclid’s Axiom
SOL – In the figure, we can see that PR coincides with
PQ + QR.
Acc. to Euclid’s Axiom, things which coincide with one another are equal to one another.
Hence Proved
NOTE – Alternative method
We can also prove using Segment addition postulate which states that if a line segment has two endpoints, A and C, a third point B lies on the line segment AC if and only if this equation AB + BC = AC
Since in figure given above, Q lies on the line segment PR Þ PR = PQ + QR.

• Step by step explanation: 
• Given:

𝑚∠COD = (x + 20)°  

𝑚∠COP = 2x°.

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

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

∠POD = ∠COD + ∠COP 

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

∠POD = ∠COD + ∠COP  

90 = (x + 20) + 2x

90 = 2x + x + 20

90 = 3x + 20

3x = 90 - 20

3x = 70

x = 

• Step 3:
• ∠COD = x + 20

 + 20

 +  

 + 

• Final Answer:

Hence, the 𝑚∠𝐶𝑂𝐷 is  .

SOLUTION:
HINT: Perform any arithmetic operation and then find.
Complete step by step solution:
Let x - y = 4…(i)
and 2x + y = 5….(ii)
On multiplying (i) with 2, we get 2( x - y = 4)
⇒ 2x - 2y = 8…(iii)
Now, we have the coefficients of  in (ii) and (iii) to be the same.
On subtracting (ii) from (iii),
we get LHS to be 2x - 2y - (2x + y) = - 2y - y = - 3y
and RHS to be 8 - 5 = 3
On equating LHS and RHS, we have - 3y = 3
⇒ y = - 1
On substituting the value of  in (i), we get - (- 1) = 4
⇒x + 1 = 4
⇒x = 4 - 1
⇒x = 3
Hence we get x = 3 and y = - 1
Note: We can also solve these system of equations by making the coefficients of y
to be the same in both the equations.

Solution :-
Hint :- find y by substituting x (in terms of y) in the equation and find x by substituting value of  y in the equations  .
Ans :- 
Explanation :-
                 — eq 1
             —- eq 2
Step 1 :- find y by substituting  in eq 2.


Y = 2
Step 2 :- substitute value of y and find x


x = 1 and y = 2  is the solution of the given pair of equations