The substitution method is one of the techniques that we use to solve a system of linear equations by expressing one variable in terms of another and substituting it into the second equation.
This method is mostly used when one equation is already solved for one variable or can be easily rearranged.
Steps
Let us solve the system of linear equations:
y = 2x + 3
3x – y = 5
Step 1: Expressing One Variable in Terms of the Other
First, we will express one variable in terms of the other variable present in the system to simplify the system.
Here, the equation (i) is already solved for y. Thus, we can substitute y = 2x + 3 into the second equation.
Step 2: Substituting Into the Second Equation
We replace the value of y in equation (ii):
3x – (2x + 3) = 5
Step 3: Solving for x
Now, we solve the above equation to get the value of x.
⇒ 3x – 2x – 3 = 5
⇒ x – 3 = 5
⇒ x = 5 + 3
⇒ x = 8
Thus, x = 8
Step 4: Finding y
Now, substituting x = 8 into equation (i), we get
y = 2x + 3
⇒ y = 2(8) + 3
⇒ y = 16 +3
⇒ y = 19
Thus, the solution is: x = 8 and y = 19
Verifying
To verify the solutions, we will substitute both values into the original equations.
From equation (i),
y = 2x + 3
⇒ 19 = 2(8) + 3
⇒ 19 = 16 + 3
⇒ 19 = 19, verified.
From equation (ii),
3x – y = 5
⇒ 3(8) – 19 = 5
⇒ 24 – 19 = 5
⇒ 5 = 5, verified.
Thus, x = 8 and y = 19 are the solutions of both equations.
Solved Examples
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Solve the system of equations by substitution:
y = x + 4
2x + y = 10
Solution:
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Given,
y = x + 4 …..(i)
2x + y = 10 …..(ii)
Step 1: Expressing One Variable in Terms of the Other
Here, the equation (i) is already solved for y.
Step 2: Substituting Into the Second Equation
Now, substituting y = x + 4 into the equation (ii),
2x + y = 10
⇒ 2x + (x + 4) = 10
Step 3: Solving for y
⇒ 2x + x + 4 = 10
⇒ 3x + 4 = 10
⇒ 3x = 10 – 4
⇒ 3x = 6
⇒ x = 2
Step 4: Finding y
Now, substituting x = 2 into equation(i), we get
y = x + 4
⇒ y = 2 + 4
⇒ y = 6
Thus, the solution is: x = 2 and y = 6
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Solve by substitution:
2x + 3y = 12
x – 4y = -5
Solution:
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Given,
2x + 3y = 12 …..(i)
x – 4y = -5 …..(ii)
Step 1: Expressing One Variable in Terms of the Other
From equation (ii),
x – 4y = -5
⇒ x = 4y – 5
Step 2: Substituting Into the Second Equation
Now, substituting x = 4y – 5 into the equation (i),
2x + 3y = 12
⇒ 2(4y – 5) + 3y = 12
Step 3: Solving for y
⇒ 8y – 10 + 3y = 12
⇒ 11y = 12 + 10
⇒ 11y = 22
⇒ y = 2
Step 4: Finding x
Now, substituting y = 2 into equation(ii), we get
x – 4y = -5
⇒ x = 4y – 5
⇒ x = 4(2) – 5
⇒ x = 8 – 5
⇒ x = 3
Thus, the solution is: x = 3 and y = 2
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Solve the system of equations:
3x – 2y = 4
x – y = 5
Solution:
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Given,
3x – 2y = 4 …..(i)
x – y = 5 …..(ii)
Step 1: Expressing One Variable in Terms of the Other
From equation (ii),
x – y = 5
⇒ x = y + 5
Step 2: Substituting Into the Second Equation
Now, substituting x = y + 5 into the equation (i),
3x – 2y = 4
⇒ 3(y + 5) – 2y = 4
Step 3: Solving for y
⇒ 3y + 15 – 2y = 4
⇒ y = 4 – 15
⇒ y = -11
Step 4: Finding x
Now, substituting y = -11 into equation(ii), we get
x – y = 5
⇒ x = y + 5
⇒ x = -11 + 5
⇒ x = -6
Thus, the solution is: x = -6 and y = -11
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Solve the given equations:
${\dfrac{1}{2}x+\dfrac{1}{3}y=5}$
x = 3y – 2
Solution:
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Given,
${\dfrac{1}{2}x+\dfrac{1}{3}y=5}$ …..(i)
x = 3y – 2 …..(ii)
Step 1: Expressing One Variable in Terms of the Other
Here, the equation (ii) is already solved for x.
Step 2: Substituting Into the Second Equation
Now, substituting x = 3y – 2 into the equation (i),
${\dfrac{1}{2}x+\dfrac{1}{3}y=5}$
⇒ ${\dfrac{1}{2}\left( 3y-2\right) +\dfrac{1}{3}y=5}$
Step 3: Solving for y
⇒ ${\dfrac{3}{2}y-1+\dfrac{1}{3}y=5}$
⇒ ${\dfrac{3}{2}y+\dfrac{1}{3}y=6}$
⇒ ${\dfrac{9y+2y}{6}=6}$
⇒ ${\dfrac{11y}{6}=6}$
⇒ 11y = 36
⇒ y = ${\dfrac{36}{11}}$
Step 4: Finding x
Now, substituting y = ${\dfrac{36}{11}}$ into equation(ii), we get
x = 3y – 2
⇒ x = ${3\left( \dfrac{36}{11}\right) -2}$
⇒ x = ${\dfrac{108-22}{11}}$
⇒ x = ${\dfrac{86}{11}}$
Thus, the solution is: x = ${\dfrac{86}{11}}$ and y = ${\dfrac{36}{11}}$
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Emma and Liam are shopping for school supplies. Emma buys 3 notebooks and 2 pens for \$8. Liam buys 5 notebooks and 3 pens for \$13. Find the price of one notebook and one pen.
Solution:
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Let x be the price of one notebook, and y be the price of one pen.
From the given problem,
Emma’s purchase is 3x + 2y = 8 …..(i)
Liam’s purchase is 5x + 3y = 13 …..(ii)
Solving for y in terms of x using the equation (i),
3x + 2y = 8
⇒ 2y = 8 – 3x
⇒ y = ${\dfrac{8-3x}{2}}$
Substituting y = ${\dfrac{8-3x}{2}}$ into the equation (ii),
${5x+3\left( \dfrac{8-3x}{2}\right) =13}$
⇒ ${\dfrac{10x+24-9x}{2}=13}$
⇒ ${\dfrac{x+24}{2}=13}$
⇒ ${x+24=26}$
⇒ ${x=26-24}$
⇒ ${x=2}$
Substituting x = 2 into the equation (i),
3x + 2y = 8
⇒ 3(2) + 2y = 8
⇒ 6 + 2y = 8
⇒ 2y = 8 – 6
⇒ 2y = 2
⇒ y = 1
Thus, one notebook costs \$2, and one pen costs \$1.
Problem: Solving for 3 VARIABLES
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Solve the following system of equations using the substitution method:
x + y – z = 1
2x – y + 3z = 9
3x + y + 2z = 8
Solution:
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Given,
x + y – z = 1 …..(i)
2x – y + 3z = 9 …..(ii)
3x + y + 2z = 8 …..(iii)
Step 1: Expressing One Variable in Terms of the Others
From equation (i),
x + y – z = 1
⇒ x = 1 – y + z …..(iv)
Step 2: Substituting x = 1 – y + z Into the Equations
Substituting (iv) into the equations (ii) and (iii),
2x – y + 3z = 9
⇒ 2(1 – y + z) – y + 3z = 9
⇒ 2 – 2y + 2z – y + 3z = 9
⇒ -3y + 5z = 7 …..(v)
3x + y + 2z = 8
⇒ 3(1 – y + z) + y + 2z = 8
⇒ 3 – 3y + 3z + y + 2z = 8
⇒ -2y + 5z = 5 …..(vi)
Step 3: Solving for y and z
Now, solving equations (v) and (vi):
-3y + 5z = 7 …..(v)
-2y + 5z = 5 …..(vi)
Step 4: Expressing z in Terms of y
From (v),
-3y + 5z = 7
⇒ 5z = 7 + 3y …..(vii)
Step 5: Substituting 5z = 7 + 3y into the Equation -2y + 5z = 5
Substituting (vii) into equation (vi),
-2y + 5z = 5
⇒ -2y + (7 + 3y) = 5
⇒ -2y + 7 + 3y = 5
⇒ -2y + 3y = 5 – 7
⇒ y = -2 …..(viii)
Step 6: Finding z
Substituting (viii) into equation (vii),
5z = 7 + 3y
⇒ 5z = 7 + 3(-2)
⇒ 5z = 7 – 6
⇒ z = ${\dfrac{1}{5}}$ …..(ix)
Step 7: Finding x
Substituting y = -2 and z = ${\dfrac{1}{5}}$ into equation (i),
x + y – z = 1
⇒ x = 1 – y + z
⇒ x = ${1-\left( -2\right) +\left( \dfrac{1}{5}\right)}$
⇒ x = ${3+\dfrac{1}{5}}$
⇒ x = ${\dfrac{16}{5}}$
Thus, x = ${\dfrac{16}{5}}$, y = -2, and z = ${\dfrac{1}{5}}$
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Solve:
x + 2y – z = 4
2x – y + 3z = 9
x + y + 2z = 13
Solution:
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Given,
x + 2y – z = 4 …..(i)
2x – y + 3z = 9 …..(ii)
x + y + 2z = 13 …..(iii)
Step 1: Expressing One Variable in Terms of the Others
From equation (i),
x + 2y – z = 4
⇒ x = 4 – 2y + z …..(iv)
Step 2: Substituting x = 4 – 2y + z Into the Equations
Substituting (iv) into the equations (ii) and (iii),
2x – y + 3z = 9
⇒ 2(4 – 2y + z) – y + 3z = 9
⇒ 8 – 4y + 2z – y + 3z = 9
⇒ -5y + 5z = 1 …..(v)
x + y + 2z = 13
⇒ (4 – 2y + z) + y + 2z = 13
⇒ 4 – 2y + z + y + 2z = 13
⇒ -y + 3z = 9 …..(vi)
Step 3: Solving for y and z
Now, solving equations (v) and (vi):
-5y + 5z = 1 …..(v)
-y + 3z = 9 …..(vi)
Step 4: Expressing y in Terms of z
From (v),
-5y + 5z = 1
⇒ 5y – 5z = -1
⇒ 5y = 5z – 1
⇒ y = ${\dfrac{5z-1}{5}}$ …..(vii)
Step 5: Substituting y = ${\dfrac{5z-1}{5}}$ into the Equation -y + 3z = 9
Substituting (vii) into equation (vi),
-y + 3z = 9
⇒ ${-\left( \dfrac{5z-1}{5}\right) +3z=9}$
⇒ ${\dfrac{-5z+1+15z}{5}=9}$
⇒ ${\dfrac{10z+1}{5}=9}$
⇒ ${10z+1=45}$
⇒ ${10z=44}$
⇒ ${z=\dfrac{22}{5}}$ …..(viii)
Step 6: Finding y
Substituting (viii) into equation (vii),
y = ${\dfrac{5z-1}{5}}$
⇒ ${y=\dfrac{5\left( \dfrac{22}{5}\right) -1}{5}}$
⇒ ${y=\dfrac{21}{5}}$ …..(ix)
Step 6: Finding x
Substituting ${y=\dfrac{21}{5}}$ and ${z=\dfrac{22}{5}}$ into equation (i),
x + 2y – z = 4
⇒ x = 4 – 2y + z
⇒ ${x=4-2\left( \dfrac{21}{5}\right) +\left( \dfrac{22}{5}\right)}$
⇒ ${x=\dfrac{20-42+22}{5}}$
⇒ ${x=0}$
Thus, x = 0, y = ${\dfrac{21}{5}}$, and z = ${\dfrac{22}{5}}$
