In this article, we will focus mainly on solving the linear equations using the first algebraic method called “Substitution Method” in detail.
Q1: Solve the equations using substitution method and find the values of a and b:
0.2a + 0.3b = 1.3
0.4a + 0.5b = 2.3.
Also, verify your answer.
Solution:
Given:
0.2a + 0.3b = 1.3 …(1)
0.4a + 0.5b = 2.3 … (2)
From equation (1), we get
0.2a = 1.3 – 0.3b
Hence, a = (1.3 – 0.3b)/0.2 …(3)
Substituting equation (3) in (2), we get
0.4[(1.3 – 0.3b)/0.2] + 0.5b = 2.3
0.4(1.3 – 0.3b) + 0.1b = 0.46
0.52 – 0.12b + 0.1b = 0.46
– 0.12b + 0.1b = 0.46 – 0.52
0.02b = 0.06
Cancelling minus sign on both sides, we get
0.02b = 0.06
Hence, b = 3.
Now, substitute b = 3 in equation (1), we get
0.4a + 0.5(3) = 2.3
0.4a + 1.5 = 2.3
0.4a = 2.3 – 1.5
0.4a = 0.8
a = 0.8 /0.4
a = 2
Hence, a = 2 and b = 3 are the solutions of the equations 0.2a + 0.3b = 1.3 and 0.4a + 0.5b = 2.3.
Verification:
Substitute a = 2 and b = 3 in equation (1).
⇒ 0.2(2) + 0.3(3) = 1.3
⇒ 0.4 + 0.9 = 1.3
⇒ 1.3 = 1.3
Thus, LHS = RHS.
Hence, a = 2 and b = 3 are the solutions to the linear equation 0.2a + 0.3b = 1.3 and 0.4a + 0.5b = 2.3.
Q2: Find the values of x and y from the system of linear equations:
7p + 6q = 3800, and 3p + 5q = 1750
Solution:
p = 500, q = 50
Given linear equations:
7p + 6q = 3800 .. (1)
3p + 5q = 1750 …(2)
From (1), we can write:
7p = 3800 – 6q
⇒ p = (3800 – 6q)/7 …(3)
Substituting (3) in (2),
⇒ 3[(3800 – 6q)/7] + 5q = 1750
⇒ 11400 – 18q + 35q = 12250
⇒ 18q + 35q = 12250 – 11400
⇒ 17q = 850
⇒ q = 850/17
⇒ q = 50
Now, plug the value of q in equation (1), we get
7p + 6(50) = 3800
7p + 300 = 3800
7p = 3800 – 300
7p = 3500
p = 3500/7
p = 500
Hence, the values of p and q are 500 and 50, respectively.
Q3: Using substitution method, solve the following equations:
p + 3q = 13 and 3p + q = 7.
Also, justify your answer.
Solution:
Given equations:
p + 3q = 13 …(1)
3p + q = 7 …(2)
From equation (1), we can write
p = 13 – 3q ..(3)
Substituting (3) in (2), we get
⇒ 3(13 – 3q) + q = 7
⇒ 39 – 9q + q = 7
⇒ 39 – 8q = 7
⇒ 8q = 7 – 39
⇒ 8q = 32
⇒ 8q = 32
⇒ q = 32/8
⇒ q = 4
Now, substituting q = 4 in equation (1), we get
⇒ p + 3(4) = 13
⇒ p + 12 = 13
⇒ p = 13 – 12
⇒ p = 1
Hence, the solutions are:
p = 1 and q = 4
Verification:
Substitute p = 1 and q = 4 in equation (1),
⇒ 1 + 3(4) = 13
⇒ 1 + 12 = 13
⇒ 13 = 13
Thus, LHS = RHS.
Therefore, p = 1 and q = 4 are the solutions of the equations p + 3q = 13 and 3p + q = 7.
Q4: Solve the pair of linear equations given below using the substitution method:
3p – q = 3
9p – 3q = 9
Solution:
Given linear equations:
3p – q = 3 …(1)
9p – 3q = 9 …(2)
From (1), we can write
⇒ 3p = 3 + q
⇒ p = (3+q)/3 …(3)
Now, substitute (3) in (2), we get
⇒ 9[(3+q)/3] – 3q = 9
⇒ 9 +3q – 3q = 9
⇒ 9 = 9
Therefore, substituting different values of q in p = (3+q)/3, we get different p values.
Thus, the system of linear equations has infinitely many solutions.
Q5: Solve the equations 2p – 3q = 2 and p + 2q = 8, using the method of substitution.
Solution:
Given:
2p – 3q = 2 …(1)
p + 2q = 8 …(2)
Using the substitution method, we can find the values of p and q as given below:
From equation (1),
⇒ 2p = 2 + 3q
⇒ p = (2 + 3q)/2
Now, substituting p = (2 + 3q)/2 in (2), we get
⇒ [(2 + 3q)/2] + 2q = 8
⇒ 2 + 3q + 4q = 16
⇒ 2 + 7q = 16
⇒7q = 16 – 2
⇒ 7q = 14
⇒ q = 14/7 = 2
Putting q = 2 in equation (2),
p + 2(2) = 8
p + 4 = 8
p = 8 – 4
p = 4
Hence, the solution to the system of equations 2p – 3q = 2 and p + 2q = 8 are:
p = 4 and q = 2.
Q6: Solve for x and y.
Equation A: y + x = 3
Equation B: x = y + 5
Solution:
The goal of the substitution method is to rewrite one of the equations in terms of a single variable. Equation B tells us that x=y+5, so it makes sense to substitute y+5 into Equation A for x.y + x = 3
x = y + 5Substitute y + 5 into Equation A for x.
y + x = 3
y + (y + 5) = 3Simplify and solve the equation for y.
Now find x by substituting this value for y into either equation and solve for x. We will use Equation A here.
Finally, check the solution x=4, y=−1 by substituting these values into each of the original equations.
Ans: x = 4 and y = −1
The solution is (4,−1).
Q7: Solve the following system of equations by substitution.
−x + y = −5
2x − 5y = 1
Solution:
First, we will solve the first equation for y.−x + y = −5
y = x − 5Now, we can substitute the expression x − 5 for y in the second equation.
2x − 5y = 1
2x − 5(x − 5) = 1
2x − 5x + 25 = 1
−3x = −24
x = 8Now, we substitute x = 8 into the first equation and solve for y.
−(8) + y = −5
y = 3Our solution is (8, 3)(8, 3).
Check the solution by substituting (8, 3)(8, 3) into both equations.
−x + y = −5
−(8) + (3) = −5 True
2x − 5y = 1
2(8) − 5(3) = 1 TrueThe substitution method can be used to solve any linear system in two variables, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions.
Q8: Solve for x and y.
2x + 3y = 22
3x + y = 19
Solution:
Choose an equation to use for the substitution. The second equation,
3x + y = 19, can easily be rewritten in terms of y, so it makes sense to start there.
2x + 3y = 22
3x + y = 19Rewrite 3x + y = 19 in terms of y.
Substitute 19 – 3x for y in the other equation.
Simplify and solve the equation for x.
Substitute x = 5 back into one of the original equations to solve for y.
Check both solutions by substituting them into each of the original equations.
Ans: x = 5 and y = 4The solution is (5, 4)(5, 4).
Q9: Solve the system of linear equations using substitution method:
p + q = 14 and p – q = 2
Solution:
Given:
p + q = 14 …(1)
p – q = 2 … (2)
From equation (1),
p = 14 – q …(3)
Now, substitute (3) in (2), we get
(14 – q) – q = 2
14 – 2q = 2
2q = 2 – 14
2q = 12
q = 12/2
q = 6
Now, substitute q = 6 in (1)
p + 6 = 14
p = 14 – 6
p = 8
Hence, the solutions are: p = 8 and q = 6.
Q10: Solve the linear equations using the substitution method:
11p + 15q = 23
7p – 2q = 20
Solution:
Given system of linear equations are:
11p + 15q = 23 …(1)
7p – 2q = 20 …(2)
From equation (1), we can write the following:
⇒ 11p = 23 – 15q
⇒ 11p = – (23 + 15q)
⇒ p = – (23 + 15q)/11 …(3)
Now, substitute equation (3) in (2)
⇒ 7[ (23 + 15q)/11] – 2q = 20
⇒ (161 + 105q) – 22q = 220
⇒ 161 – 105q – 22q = 220
⇒ 105q – 22q = 220 + 161
⇒ – 127q = 381
⇒ q = 381/127
⇒ q = 3
Substituting q = 3 in (2),
⇒ 7p – 2(3) = 20
⇒ 7p + 6 = 20
⇒ 7p = 20 – 6
⇒ 7p = 14
⇒ p = 14/7
⇒ p = 2
Therefore, the values of p and q are 2 and 3, respectively.
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