Tenth Grade Mathematics: Algebraic Solution: Method of Substitution

While graphical methods provide geometric intuition, algebraic techniques deliver exact, unambiguous solutions without relying on grid accuracy. The first of the primary algebraic techniques is the Method of Substitution. Based on the fundamental algebraic principle of replacement, this method expresses one variable explicitly in terms of the other from one equation, and then substitutes that expression into the second equation, collapsing a two-variable system into a single solvable linear equation.

Algorithmic Procedure of the Substitution Method

The method follows four sequential, systematic steps:

Step 1: Choose one equation and isolate one variable (say y in terms of x, or
        x in terms of y). Select the variable with a coefficient of 1 or -1
        whenever possible to avoid fractional arithmetic.

Step 2: Substitute this isolated expression into the OTHER equation.
        This eliminates that variable completely, yielding a single linear
        equation in one unknown.

Step 3: Solve this single-variable linear equation for the remaining unknown.

Step 4: Substitute the obtained numerical value back into the expression from
        Step 1 to compute the exact value of the first variable.

Visual Workflow:
Equation 1: [ a1*x + b1*y + c1 = 0 ] =======> Express y = f(x)
                                                    |
                                                    v Substitute into Equation 2
Equation 2: [ a2*x + b2*y + c2 = 0 ] =======> a2*x + b2*[f(x)] + c2 = 0
                                                    |
                                                    v Solve for x
                                              x = Value
                                                    |
                                                    v Substitute into y = f(x)
                                              y = Value

Detailed Worked Examples

Example 1: Standard Integer System

Solve the system of equations using substitution:

Equation 1:  7*x - 15*y = 2
Equation 2:  x + 2*y = 3

Step 1: In Equation 2, the variable x has a coefficient of 1, making it the most efficient choice to isolate:

x = 3 - 2*y      --- (Equation 3)

Step 2: Substitute Equation 3 into Equation 1:

7*(3 - 2*y) - 15*y = 2
21 - 14*y - 15*y = 2
21 - 29*y = 2

Step 3: Solve for y:

-29*y = 2 - 21
-29*y = -19
y = 19 / 29

Step 4: Substitute y = 19/29 back into Equation 3 to find x:

x = 3 - 2*(19 / 29)
x = 3 - 38 / 29 = (87 - 38) / 29 = 49 / 29

The unique solution is x = 49/29, y = 19/29.

Example 2: Recognizing Special Cases via Substitution

When solving a system that is either inconsistent (parallel lines) or dependent (coincident lines), the substitution method automatically reveals the structural reality:

Case A: Inconsistent System
Equation 1:  x + 2*y = 4    ===>  x = 4 - 2*y
Equation 2:  2*x + 4*y = 12
Substitute x into Equation 2:
2*(4 - 2*y) + 4*y = 12
8 - 4*y + 4*y = 12
8 = 12   (A mathematically FALSE statement!)
Conclusion: Because 8 = 12 is impossible, there is NO SOLUTION. The system is inconsistent.

Case B: Dependent System
Equation 1:  2*x + 3*y = 9   ===>  x = (9 - 3*y) / 2
Equation 2:  4*x + 6*y = 18
Substitute x into Equation 2:
4*((9 - 3*y) / 2) + 6*y = 18
2*(9 - 3*y) + 6*y = 18
18 - 6*y + 6*y = 18
18 = 18  (A mathematically TRUE identity for all y!)
Conclusion: The equations are dependent; there are INFINITELY MANY SOLUTIONS.

Chapter Exercises

Exercise 1: Solve the following pair of linear equations by the method of substitution: s - t = 3 and s/3 + t/2 = 6.

Exercise 2: Solve the system: 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 using substitution. Clear all decimals first.

Exercise 3: Solve the radical linear system: sqrt(2)x + sqrt(3)y = 0 and sqrt(3)x - sqrt(8)y = 0 by the method of substitution.

Solutions and Explanations

Solution 1: From the first equation s - t = 3, express s in terms of t: s = t + 3. Clear the fractions in the second equation s/3 + t/2 = 6 by multiplying by the LCM of 3 and 2, which is 6: 6(s/3) + 6(t/2) = 6(6), giving 2s + 3t = 36. Now substitute s = t + 3 into this equation: 2(t + 3) + 3t = 36 ==> 2t + 6 + 3t = 36 ==> 5t + 6 = 36 ==> 5*t = 30 ==> t = 6. Substitute t = 6 back into the expression for s: s = 6 + 3 = 9. Thus, the solution is s = 9, t = 6.

Solution 2: Multiply both equations by 10 to clear decimals: Equation 1 becomes 2x + 3y = 13, and Equation 2 becomes 4x + 5y = 23. From Equation 1, express x in terms of y: 2x = 13 - 3y ==> x = (13 - 3y)/2. Substitute this expression for x into Equation 2: 4((13 - 3y)/2) + 5y = 23 ==> 2(13 - 3y) + 5y = 23 ==> 26 - 6y + 5y = 23 ==> 26 - y = 23 ==> -y = 23 - 26 = -3 ==> y = 3. Substitute y = 3 back into the expression for x: x = (13 - 3(3))/2 = (13 - 9)/2 = 4/2 = 2. Hence, the solution is x = 2, y = 3.

Solution 3: From Equation 1: sqrt(2)x + sqrt(3)y = 0, express x in terms of y: sqrt(2)x = -sqrt(3)y ==> x = -(sqrt(3)/sqrt(2))y. Substitute this into Equation 2: sqrt(3)x - sqrt(8)y = 0 ==> sqrt(3)[-(sqrt(3)/sqrt(2))y] - sqrt(8)y = 0 ==> -(3/sqrt(2))y - sqrt(8)y = 0. Factor out y: y * [-(3/sqrt(2)) - 2sqrt(2)] = 0. Notice that the expression in the bracket is strictly negative and non-zero (-(3/sqrt(2)) - 4/sqrt(2) = -7/sqrt(2) != 0). Dividing by this non-zero bracket yields y = 0. Substituting y = 0 back into x gives x = -(sqrt(3)/sqrt(2))(0) = 0. Thus, the unique solution is x = 0, y = 0.