1.4 Solving by Substitution
To solve a system of linear equations by using substitution includes the following steps.
Consider we have system of equations as:
\(x + y = 10\) and \(2x + 3y = 8\)
Step 1: Isolate one of the variables in one of the equations
Choose one of the equations (any of the variables having coefficient as ‘1’) and solve it for one variable in terms of the other. This can be either x or y. For example, if you have the equation \(x + y = 10\), you can isolate \(x\) by subtracting \(y\) from both sides to get \(x = 10 - y\).
Step 2: Substitute the expression from Step 1 into the other equation
Substitute the expression you found in Step 1 into the other equation. This will give you an equation with only one variable. For example, if the second equation is \(2x + 3y = 8\) and \(x = 10 - y\), substitute \(x\) in the second equation to get \(2(10 - y) + 3y = 8\).
Step 3: Solve the new equation
Solve the equation from Step 2. This will give you the value of one of the variables. In our example, solving \(2(10 - y) + 3y = 8\) gives \(y = -12\).
Step 4: Substitute the result from Step 3 into either of the original equations
Substitute the value you found in Step 3 back into either of the original equations to find the value of the other variable. In our example, substituting \(y = -12\) into the first equation \(x + y = 10\) gives \(x = 10 -(-12) = 22\). So, the solution to the system of equations is \(x = 22, y = -12\).
Step 5: Check the solution
Substitute the values of \(x\) and \(y\) into both original equations to verify that they are true. If both equations are true with the solution, then you have solved the system of equations correctly.
Check: plug \(x=22\) and \(y=-12\) in equation (1)
LHS: \(22-12=10\)=RHS
plug \(x=22\) and \(y=-12\) in equation (2)
LHS: \(2(22)+3(-12)=44-36\)
\(=8=\)RHS
So, values of \(x\) and \(y\) are true for both the equations.