Solving Two Equations by Substitution

From Two Equations to One

A solution to a two-variable system must satisfy both equations at once. Substitution uses one equation to express one variable in terms of the other, then replaces that variable in the second equation. The result is a single equation with one unknown. After solving it, substitute back to find the other coordinate. The method is useful because each step follows from equality: the replacement does not change which ordered pairs satisfy the system.

Unless stated otherwise, the variables and coefficients in this tutorial are real numbers. We write an ordered pair as \((x,y)\), so its first coordinate is the value of \(x\) and its second coordinate is the value of \(y\). A system such as \(a x+b y=c\) and \(d x+e y=f\) asks for all ordered pairs that satisfy both equations. The substitution steps below focus on finding a pair when the resulting one-variable equation determines one.

For example, if \(2x+y=7\), then \(y=7-2x\). Substituting \(7-2x\) for \(y\) in another equation is valid because the first equation says that \(y\) and \(7-2x\) have the same value for every solution of the system. In contrast, dividing an equation by a coefficient that is zero is never valid. When choosing what to isolate, check that the coefficient of that variable is nonzero.

Why Substitution Preserves the Solutions

Proof. Since \(a\ne0\), the equation \(a x+b y=c\) can be rearranged by subtracting \(b y\) from both sides and dividing both sides by \(a\), giving \(x=\frac{c-by}{a}\). If \((x,y)\) solves the original system, this formula holds, and replacing \(x\) by \(\frac{c-by}{a}\) in the second equation gives \(d\left(\frac{c-by}{a}\right)+e y=f\). Conversely, suppose \(y\) satisfies that one-variable equation and set \(x=\frac{c-by}{a}\). Then \(a x+b y=a\left(\frac{c-by}{a}\right)+b y=c-by+by=c\), and the assumed one-variable equation gives \(d x+e y=f\). Thus the resulting pair satisfies both original equations. These two implications show that the substitution step neither loses nor adds solutions. \(\square\)

This proposition explains the basic workflow: the first equation supplies a formula, and the second equation supplies the condition on the remaining variable. If the coefficient of \(y\) in the first equation is nonzero, the same argument works with \(y\) isolated instead. The useful choice is often the one that gives a short expression and avoids fractions, though fractions are perfectly valid when needed.

Worked Examples

Choosing an Equation and Reading the Result

Either equation can be used to isolate a variable, as long as the coefficient being divided by is nonzero. The result is equivalent whichever equation is chosen, but the amount of arithmetic may differ. For instance, if one equation already reads \(y=4-x\), use that expression directly rather than rearranging the other equation first. If an equation has a coefficient of \(1\) or \(-1\), isolating the corresponding variable often keeps the calculations especially simple.

Proof. If \(a\ne0\), subtract \(b y\) from \(a x+b y=c\) to obtain \(a x=c-by\), then divide both sides by \(a\) to obtain \(x=\frac{c-by}{a}\). Reversing these steps is allowed because \(a\ne0\): multiplying by \(a\) and adding \(b y\) recovers \(a x+b y=c\). If \(b\ne0\), subtract \(a x\) to obtain \(b y=c-ax\), then divide by \(b\); reversing those steps recovers the original equation. Therefore each displayed form is equivalent to its original equation under the stated condition. \(\square\)

After substitution, the remaining equation may have a nonzero coefficient on its variable, so that variable can be found by division. If its coefficient is zero, do not divide by zero. Instead, read the resulting statement as it stands: it may be a false numerical equality, or it may hold for every value of the remaining variable. In either situation, the calculation has not produced one uniquely determined value by ordinary division. Detailed descriptions of those system outcomes belong with the study of solution sets.

Common Mistakes

Sign error when substituting. Consider \(x+2y=8\), \(3x-y=3\). The first equation gives \(x=8-2y\). Substituting into the second must keep the minus sign before \(y\). A wrong calculation is

The term \(-y\) remains negative, so the correct expansion is \(24-6y-y=3\). Hence \(-7y=-21\), \(y=3\), and \(x=8-2(3)=2\). Check: \(2+2(3)=8\), and \(3(2)-3=3\). The wrong expansion would instead give \(y=\frac{21}{5}\) and \(x=-\frac{2}{5}\), which fails the second equation because \(3\left(-\frac{2}{5}\right)-\frac{21}{5}=-\frac{27}{5}\ne3\).

Other checks to make:

• Use parentheses when substituting an expression such as \(5x-7\), especially if it is multiplied by a coefficient.
• Do not divide by a zero coefficient. Try isolating the other variable or use an equation with a nonzero coefficient.
• Do not reverse the coordinate order. If \(x=3\) and \(y=1\), the ordered pair is \((3,1)\), not \((1,3)\).
• Check the final pair in both original equations. A value obtained from one equation alone need not satisfy the other.

Substitution is more than a shortcut for arithmetic: it is a sequence of equivalent replacements and reversible algebraic steps. Keeping track of why each step is valid helps prevent sign errors and makes the final check meaningful. It also gives a reliable method for solving small systems when one equation can be rearranged conveniently.