Solve by the elimination method of these 2 questions..?

1)
3x - 2y = 13
5x + 6y = 3

3(3x - 2y) = 3(13)
9x - 6y = 39

...9x - 6y = 39
+) 5x + 6y = 3 (addition)
-----------------------------
14x = 42

14x = 42
x = 42/14
x = 3

3x - 2y = 13
3(3) - 2y = 13
9 - 2y = 13
2y = 9 - 13
2y = -4
y = -4/2
y = -2

∴ x = 3 , y = -2

= = = = = = = =

2)
p = 31 + q
5p = -3q + 211

p = 31 + q
5(p) = 5(31 + q)
5p = 155 + 5q
5p = 5q + 155

...5p = 5q + 155
–) 5p = -3q + 211 (subtraction)
-------------------------------
0 = 8q - 56

0 = 8q - 56
8q = 56
q = 56/8
q = 7

p = 31 + q
p = 31 + 7
p = 38

∴ p = 38 , q = 7

Q1:
Value of x:
(3x - 13)/2 = (3 - 5x)/6
3(3x - 13) = 3 - 5x
9x - 39 = 3 - 5x
14x = 42
x = 3

Value of y (1st equation):
= (3[3] - 13)/2
= (9 - 13)/2
= - 4/2 or - 2

Answer: x = 3, y = - 2

Proof (2nd equation):
5(3) + 6(- 2) = 3
15 - 12 = 3
3 = 3

Q2:
Value of q:
31 + q = (- 3q + 211)/5
155 + 5q = - 3q + 211
8q = 56
q = 7

Value of p (1st equation):
= 31 + 7
= 38

Answer: p = 38, q = 7

Proof (2nd equation):
5(38) = - 3(7) + 211
190 = - 21 + 211
190 = 190

9x - 6y = 39
5x + 6y = 3-------ADD

14x = 42
x = 3

15 + 6y = 3
6y = - 12
y = - 2

x = 3 , y = - 2

5(31 + q) = - 3q + 211
155 + 5q = - 3q + 211
8q = 56
q = 7

p + 31 + t
p = 38

p = 38 , q = 7

3x-2y=13 (A)
5x+6y=3 (B)
Multiply equation (A) by 3 to get
9x-6y=39 (C)


9x-6y=39 (C)
5x+6y=3 (B)
Add equations (C) & (B) together to get
14x =42
Divide both sides by 14 to get
x =3

Substitute x=3 into equation (B) to get
15+6y=3
Subtract 15 from both sides to get
6y=-12
Divide both sides by 6 to get
y=-2

Solution x=3 , y=-2
Check your answer by substitution!

Put x=3 & y= -2 into equation (A)
LHS=3.3-2.(-2)=9+4=13=RHS correct!

Put x=3 & y= -2 into equation (B)
LHS=5.3+6.(-2)=15-12=3=RHS correct!
____________________________________

p=31+q (A)
5p=-3q+211 (B)
The easiest way to solve this would be by the substitution method.

From equation (A) p=31+q
Substitute in equation (B) to get
5(31+q)= -3q+211

Multiply out the bracket to get
155+5q= -3q+211

Subtract 155 from both sides to get
5q= -3q+56

Add 3q to each side to get
8q=56

Divide both sides by 8 to get
q=7

Substitute q=7 in equation (A) to get
p=31+7=38

Solution is p=38 & q=7
Check your answer by substitution!

Substitute p=38 & q=7 in equation (A)
RHS=31+7=38=LHS correct!
Substitute p=38 & q=7 in equation (B)
RHS= -3.7+211=190 LHS=5.38=190 correct!

1) 3x - 2y = 13
2) 5x + 6y = 3
3) # multiply 1) with 3, add 2)
9x - 6y = 39
5x + 6y = 3
-----------------
4) 14x = 42
5) x = 3
6) 3*3 - 2y = 13
7) 4 = 2y
8) y= 2
S{ 3; 2 }
--2nd--
1) p = 31 + q
2) 5p =- 3q + 211
3) 5(31 + q) = -3q + 211
4) 155 + 5q = -3q + 211
5) 8q = 56
9) q = 7
10) p = 31 + 7 = 38
S{ 38; 7 }

For the first one,
multiply the first equation by 3 so you can cancel out the y values.
9x - 6y = 39
5x +6y = 3
14x = 42
x = 3
then plug it in to one of the original equations
3(3) - 2y = 13
9 - 2y = 13
-2y = 4
y = -2


for the second problem
just plug p into the second equation, use the distributive property and solve for q.

5(31+q) = -3q+211
155 + 5q = -3q + 211
155 + 8q = 211
8q = 56
q = 7
then plug q in for the equation to get p,
p = 31 + 7
p = 38