Solving Systems of Equations by Substitution

The substitution method consists of isolating one variable in one of the equations of a system and substituting its algebraic expression into the other equation. Through this procedure, a system of two linear equations in two variables (2×2) is transformed into a single linear equation in one variable, which can be solved directly.

This method is particularly efficient when at least one of the variables has a coefficient of 1 or −1, as it allows isolating the variable without introducing fractions in the initial steps. Although primarily applied to 2×2 systems, this algebraic principle can also be extended to systems of three linear equations in three variables (3×3) by eliminating one variable at a time until the system is reduced.

Example

To illustrate the procedure on a 2×2 system, consider the following introductory example:

$$\begin{cases} x + 3y = 6 \\ 5x - 2y = 13 \end{cases}$$

We label each equation with a number to maintain clarity throughout the solution process:

$$x + 3y = 6 \quad (1)$$

$$5x - 2y = 13 \quad (2)$$

We examine the coefficients of the variables in both equations. In equation (1), the variable x has a coefficient of 1, making it the most direct choice to solve for without introducing fractions:

$$x = 6 - 3y \quad (3)$$

Next, we substitute the expression found for x into equation (2). Enclosing the expression in parentheses is essential to apply the distributive property correctly:

$$5(6 - 3y) - 2y = 13$$

Multiply through the parentheses and combine like terms containing the variable y:

$$30 - 15y - 2y = 13$$

$$30 - 17y = 13$$

Subtract 30 from both sides and divide by −17 to obtain the numerical value of the first variable:

$$-17y = 13 - 30$$

$$-17y = -17$$

$$y = \frac{-17}{-17} = 1$$

With \(y = 1\) determined, substitute this value back into equation (3), where x is already isolated:

$$x = 6 - 3(1)$$

$$x = 6 - 3 = 3$$

Thus, the solution to the system is \(x = 3\) and \(y = 1\). In the Cartesian coordinate plane, this solution represents the point of intersection of the two lines and is typically written as the ordered pair \((x, y) = (3, 1)\), or as the solution set \(S = \{(3, 1)\}\).

To verify that these values simultaneously satisfy both equations, substitute \(x = 3\) and \(y = 1\) into the original equations (1) and (2):

In equation (1):

$$3 + 3(1) = 3 + 3 = 6$$

In equation (2):

$$5(3) - 2(1) = 15 - 2 = 13$$

Both statements are true, confirming that the solution is correct.

Steps to Solve a System by Substitution

To solve any system of two linear equations in two variables using this method, follow this systematic sequence of steps:

1. Isolate one variable: Choose one of the two equations and solve for one variable, prioritizing a variable with a coefficient of 1 or −1 to avoid working with fractions.
2. Substitute into the other equation: Replace that variable in the remaining equation with the resulting expression, placing the expression inside parentheses.
3. Solve the resulting linear equation: Apply algebraic operations (the distributive property and combining like terms) to solve for the remaining variable.
4. Find the other variable: Substitute the value found in the previous step into the equation isolated in step 1 to determine the value of the second variable.
5. Check the solution: Substitute both numerical values into the original equations to verify that both statements hold true.

Types of Solutions

When applying the substitution method to a system of two linear equations in two variables, the outcome is not always a single numerical value for each variable. Depending on the geometric relationship between the lines, three distinct algebraic scenarios can occur:

System Classification	Algebraic Result After Substitution	Number of Solutions	Geometric Interpretation
Consistent and independent	A unique numerical value is obtained for each variable: \(x = a\), \(y = b\).	Exactly one solution \((a, b)\).	Two intersecting lines that cross at a single point.
Consistent and dependent	The variables cancel out completely, producing a true identity such as \(0 = 0\) or \(c = c\).	Infinitely many solutions.	Coincident lines (the equations represent the exact same line).
Inconsistent	The variables cancel out completely, producing a false contradiction such as \(0 = k\) where \(k \neq 0\).	No solution.	Parallel lines that never intersect.

To quickly identify these special cases during algebraic work, observe how the equations behave when the variables drop out:

• True identity (0 = 0): If substituting and simplifying eliminates all variable terms and leaves an identity where both sides are equal, every point on the first line also lies on the second line. The equations are dependent, and the system has infinitely many solutions.
• Contradiction (0 = k): If the variables cancel out and yield a false statement (such as 0 = 8 or -5 = 3), the system represents a condition that cannot be satisfied. Geometrically, the lines are parallel and distinct (they share the same slope but have different intercepts), meaning they never intersect.

Step-by-Step Examples

Below are detailed examples arranged in order of increasing algebraic complexity. In each case, the equations are numbered to clearly follow the isolating steps, substitutions, and checks. The final examples cover practical word problems.

Example 1

Solve the following system of linear equations:

$$\begin{cases} x + 2y = 7 \\ 3x + 4y = 17 \end{cases}$$

Solution

Label each equation in the system:

$$x + 2y = 7 \quad (1)$$

$$3x + 4y = 17 \quad (2)$$

Solve for x in equation (1), since its coefficient is 1:

$$x = 7 - 2y \quad (3)$$

Substitute the expression for x into equation (2):

$$3(7 - 2y) + 4y = 17$$

Apply the distributive property and combine like terms containing y:

$$21 - 6y + 4y = 17$$

$$21 - 2y = 17$$

Subtract 21 from both sides and solve for the variable:

$$-2y = 17 - 21$$

$$-2y = -4$$

$$y = \frac{-4}{-2} = 2$$

Substitute \(y = 2\) into equation (3) to find x:

$$x = 7 - 2(2)$$

$$x = 7 - 4 = 3$$

The solution to the system is \(x = 3\) and \(y = 2\), written as the ordered pair \((x, y) = (3, 2)\).

Check the result by substituting into the original equations:

In (1):

$$3 + 2(2) = 3 + 4 = 7$$

In (2):

$$3(3) + 4(2) = 9 + 8 = 17$$

Both equations are satisfied.

Example 2

$$\begin{cases} 2x + y = 7 \\ 3x + 2y = 12 \end{cases}$$

Solution

Assign a reference number to each equation:

$$2x + y = 7 \quad (1)$$

$$3x + 2y = 12 \quad (2)$$

In this system, the variable y in equation (1) has a coefficient of 1. Solve for y in (1):

$$y = 7 - 2x \quad (3)$$

Substitute this expression into equation (2):

$$3x + 2(7 - 2x) = 12$$

Distribute the factor of 2 and combine like terms:

$$3x + 14 - 4x = 12$$

$$-x + 14 = 12$$

Isolate the variable x:

$$-x = 12 - 14$$

$$-x = -2$$

$$x = 2$$

Determine the value of y by substituting \(x = 2\) into equation (3):

$$y = 7 - 2(2)$$

$$y = 7 - 4 = 3$$

The solution is \(x = 2\) and \(y = 3\), corresponding to the point \((x, y) = (2, 3)\).

Check the values in both original equations:

In (1):

$$2(2) + 3 = 4 + 3 = 7$$

In (2):

$$3(2) + 2(3) = 6 + 6 = 12$$

Both equations are verified.

Example 3

$$\begin{cases} 2x + 3y = 8 \\ 4x - y = 2 \end{cases}$$

Solution

Number the equations in the system:

$$2x + 3y = 8 \quad (1)$$

$$4x - y = 2 \quad (2)$$

Notice that in equation (2), the variable y has a coefficient of −1. To isolate it directly without dealing with negative signs at the end, add y and subtract 2 on both sides:

$$4x - 2 = y \implies y = 4x - 2 \quad (3)$$

Substitute this expression into equation (1):

$$2x + 3(4x - 2) = 8$$

Apply the distributive property:

$$2x + 12x - 6 = 8$$

$$14x - 6 = 8$$

Add 6 to both sides and solve for x:

$$14x = 8 + 6$$

$$14x = 14$$

$$x = \frac{14}{14} = 1$$

Substitute \(x = 1\) into equation (3) to find the value of y:

$$y = 4(1) - 2$$

$$y = 4 - 2 = 2$$

The solution to the system is the ordered pair \((x, y) = (1, 2)\).

Check the ordered pair in the original equations:

In (1):

$$2(1) + 3(2) = 2 + 6 = 8$$

In (2):

$$4(1) - 2 = 4 - 2 = 2$$

The system is verified.

Example 4

$$\begin{cases} -x + 3y = 8 \\ 2x - 5y = -13 \end{cases}$$

Solution

Write the equation references:

$$-x + 3y = 8 \quad (1)$$

$$2x - 5y = -13 \quad (2)$$

In equation (1), the variable x has a coefficient of −1. Add x to both sides and subtract 8 to isolate it:

$$3y - 8 = x \implies x = 3y - 8 \quad (3)$$

Substitute this expression into equation (2):

$$2(3y - 8) - 5y = -13$$

Distribute the factor of 2 carefully with respect to signs, and combine like terms:

$$6y - 16 - 5y = -13$$

$$y - 16 = -13$$

Add 16 to both sides to solve for y:

$$y = -13 + 16$$

$$y = 3$$

Substitute \(y = 3\) into equation (3) to compute x:

$$x = 3(3) - 8$$

$$x = 9 - 8 = 1$$

The solution is \(x = 1\) and \(y = 3\), which corresponds to the point \((x, y) = (1, 3)\).

Check in the initial equations:

In (1):

$$-(1) + 3(3) = -1 + 9 = 8$$

In (2):

$$2(1) - 5(3) = 2 - 15 = -13$$

Both statements confirm the solution.

Example 5

Solve and classify the following system of linear equations:

$$\begin{cases} 3x - y = 3 \\ 9x - 3y = 9 \end{cases}$$

Solution

Identify the equations:

$$3x - y = 3 \quad (1)$$

$$9x - 3y = 9 \quad (2)$$

Isolate the variable y in equation (1):

$$3x - 3 = y \implies y = 3x - 3 \quad (3)$$

Substitute the expression for y into equation (2):

$$9x - 3(3x - 3) = 9$$

Apply the distributive property with the factor −3:

$$9x - 9x + 9 = 9$$

The variable terms cancel out:

$$0x + 9 = 9$$

$$9 = 9$$

Equivalently, subtracting 9 from both sides yields \(0 = 0\).

Because the variables drop out and produce a true identity, the statement holds for all real numbers. This indicates that the two equations are equivalent and represent the same line in the coordinate plane.

Therefore, the system is consistent and dependent and has infinitely many solutions. In set-builder notation, the solution set consists of all ordered pairs of the form \(\{(x, y) \mid y = 3x - 3\}\) or \((x, 3x - 3)\) for any \(x \in \mathbb{R}\).

Example 6

Find the solution to the system:

$$\begin{cases} 2x + 3y = 1 \\ 3x + 4y = 0 \end{cases}$$

Solution

Number the equations in the system:

$$2x + 3y = 1 \quad (1)$$

$$3x + 4y = 0 \quad (2)$$

Since no variable has a coefficient of 1 or −1, isolating any variable will necessarily introduce a fraction. Solve for x in equation (1):

$$2x = 1 - 3y \implies x = \frac{1 - 3y}{2} \quad (3)$$

Substitute this fractional expression into equation (2):

$$3\left(\frac{1 - 3y}{2}\right) + 4y = 0$$

Multiply the numerator by 3:

$$\frac{3 - 9y}{2} + 4y = 0$$

To eliminate the denominator 2, multiply every term in the equation by 2:

$$(3 - 9y) + 2(4y) = 2(0)$$

$$3 - 9y + 8y = 0$$

Combine like terms and solve for y:

$$3 - y = 0 \implies y = 3$$

Find x by substituting \(y = 3\) into equation (3):

$$x = \frac{1 - 3(3)}{2} = \frac{1 - 9}{2} = \frac{-8}{2} = -4$$

The solution to the system is \(x = -4\) and \(y = 3\), which corresponds to the point \((x, y) = (-4, 3)\).

Check the result in the original equations:

In (1):

$$2(-4) + 3(3) = -8 + 9 = 1$$

In (2):

$$3(-4) + 4(3) = -12 + 12 = 0$$

Both equations are satisfied simultaneously.

Example 7

$$\begin{cases} 2x - y = 5 \\ 4x - 2y = 7 \end{cases}$$

Solution

Identify the two equations:

$$2x - y = 5 \quad (1)$$

$$4x - 2y = 7 \quad (2)$$

Isolate the variable y in equation (1) by adding y and subtracting 5:

$$2x - 5 = y \implies y = 2x - 5 \quad (3)$$

Substitute this expression into equation (2):

$$4x - 2(2x - 5) = 7$$

Apply the distributive property with the factor −2:

$$4x - 4x + 10 = 7$$

Subtracting the variable terms eliminates x:

$$0x + 10 = 7$$

$$10 = 7$$

Equivalently, subtracting 10 from both sides gives the false statement \(0 = -3\).

Because the variables drop out completely and leave a false arithmetic contradiction, no ordered pair \((x, y)\) can satisfy both equations simultaneously. In the coordinate plane, the lines have identical slopes but different y-intercepts, meaning they are parallel lines that never intersect.

Therefore, the system is inconsistent and has no solution. Its solution set is the empty set: \(S = \emptyset\).

Example 8

Determine the solution to the following linear system:

$$\begin{cases} 2x + 5y = 5 \\ 3x + 7y = 36 \end{cases}$$

Solution

Assign a reference number to each equation:

$$2x + 5y = 5 \quad (1)$$

$$3x + 7y = 36 \quad (2)$$

Solve for x in equation (1):

$$2x = 5 - 5y \implies x = \frac{5 - 5y}{2} \quad (3)$$

Substitute this expression into equation (2):

$$3\left(\frac{5 - 5y}{2}\right) + 7y = 36$$

Distribute the factor of 3 in the numerator:

$$\frac{15 - 15y}{2} + 7y = 36$$

Multiply the entire equation by 2 to clear fractions:

$$(15 - 15y) + 2(7y) = 2(36)$$

$$15 - 15y + 14y = 72$$

Combine like terms and group the constants:

$$15 - y = 72$$

$$-y = 72 - 15$$

$$-y = 57 \implies y = -57$$

Substitute \(y = -57\) into equation (3):

$$x = \frac{5 - 5(-57)}{2} = \frac{5 + 285}{2} = \frac{290}{2} = 145$$

The solution is \(x = 145\) and \(y = -57\), written as the ordered pair \((x, y) = (145, -57)\).

Verify the solution in both original equations:

In (1):

$$2(145) + 5(-57) = 290 - 285 = 5$$

In (2):

$$3(145) + 7(-57) = 435 - 399 = 36$$

Both statements are true.

Example 9

Find the values of x and y that satisfy the system:

$$\begin{cases} \dfrac{x}{2} + \dfrac{y}{3} = 4 \\ \\ x - \dfrac{y}{2} = 1 \end{cases}$$

Solution

Identify the initial equations:

$$\frac{x}{2} + \frac{y}{3} = 4 \quad (1)$$

$$x - \frac{y}{2} = 1 \quad (2)$$

When a system contains fractions, the most efficient approach is to clear denominators by multiplying each equation by the least common multiple (LCM) of its denominators, producing an equivalent system with integer coefficients.

In equation (1), the denominators are 2 and 3, whose LCM is 6. Multiply both sides by 6:

$$6\left(\frac{x}{2}\right) + 6\left(\frac{y}{3}\right) = 6(4) \implies 3x + 2y = 24 \quad (1')$$

In equation (2), multiply both sides by 2 to clear the denominator:

$$2(x) - 2\left(\frac{y}{2}\right) = 2(1) \implies 2x - y = 2 \quad (2')$$

Now working with equations with integer coefficients, isolate the variable y in equation (2'):

$$2x - 2 = y \implies y = 2x - 2 \quad (3)$$

Substitute this expression into equation (1'):

$$3x + 2(2x - 2) = 24$$

Apply the distributive property and combine like terms in x:

$$3x + 4x - 4 = 24$$

$$7x - 4 = 24$$

Add 4 to both sides and divide by 7:

$$7x = 24 + 4$$

$$7x = 28$$

$$x = \frac{28}{7} = 4$$

Calculate y by substituting \(x = 4\) into equation (3):

$$y = 2(4) - 2 = 8 - 2 = 6$$

The point of intersection is \((x, y) = (4, 6)\).

Check the values directly in the original fractional equations:

In (1):

$$\frac{4}{2} + \frac{6}{3} = 2 + 2 = 4$$

In (2):

$$4 - \frac{6}{2} = 4 - 3 = 1$$

The solution is fully verified.

Example 10

$$\begin{cases} \dfrac{2x + y}{3} = 5 \\ \\ \dfrac{x - 3y}{4} = 2 \end{cases}$$

Solution

Number the given equations:

$$\frac{2x + y}{3} = 5 \quad (1)$$

$$\frac{x - 3y}{4} = 2 \quad (2)$$

When an entire side of an equation is divided by a constant, clear the fraction by multiplying both sides by that denominator.

In equation (1), multiply both sides by 3:

$$2x + y = 5 \cdot 3 \implies 2x + y = 15 \quad (1')$$

In equation (2), multiply both sides by 4:

$$x - 3y = 2 \cdot 4 \implies x - 3y = 8 \quad (2')$$

Isolate the variable x in equation (2') by adding \(3y\):

$$x = 8 + 3y \quad (3)$$

Substitute the expression for x into equation (1'):

$$2(8 + 3y) + y = 15$$

Distribute the factor of 2 and combine like terms with y:

$$16 + 6y + y = 15$$

$$16 + 7y = 15$$

Subtract 16 from both sides:

$$7y = 15 - 16$$

$$7y = -1$$

$$y = -\frac{1}{7}$$

Find the value of x by substituting \(y = -\frac{1}{7}\) into equation (3):

$$x = 8 + 3\left(-\frac{1}{7}\right) = 8 - \frac{3}{7}$$

Write 8 with a common denominator of 7 by multiplying by \(\frac{7}{7}\):

$$x = \frac{56}{7} - \frac{3}{7} = \frac{53}{7}$$

The solution is given by the exact fractions \(x = \frac{53}{7}\) and \(y = -\frac{1}{7}\), written as the ordered pair \((x, y) = \left(\frac{53}{7}, -\frac{1}{7}\right)\).

Check the results by substituting into the original expressions:

In (1):

$$\frac{2\left(\frac{53}{7}\right) + \left(-\frac{1}{7}\right)}{3} = \frac{\frac{106}{7} - \frac{1}{7}}{3} = \frac{\frac{105}{7}}{3} = \frac{15}{3} = 5$$

In (2):

$$\frac{\frac{53}{7} - 3\left(-\frac{1}{7}\right)}{4} = \frac{\frac{53}{7} + \frac{3}{7}}{4} = \frac{\frac{56}{7}}{4} = \frac{8}{4} = 2$$

Both equations confirm the solution.

Example 11

Find the solution to the system:

$$\begin{cases} y = 2x - 3 \\ 3x + 2y = 8 \end{cases}$$

Solution

Identify the two given equations:

$$y = 2x - 3 \quad (1)$$

$$3x + 2y = 8 \quad (2)$$

When one of the equations already has a variable isolated on one side, step 1 of the method is already complete. There is no need to rearrange the terms; substitute the expression for y from equation (1) directly into equation (2):

$$3x + 2(2x - 3) = 8$$

Apply the distributive property with the factor 2:

$$3x + 4x - 6 = 8$$

$$7x - 6 = 8$$

Add 6 to both sides and solve for the variable x:

$$7x = 8 + 6$$

$$7x = 14$$

$$x = \frac{14}{7} = 2$$

Find the value of y by substituting \(x = 2\) into equation (1):

$$y = 2(2) - 3 = 4 - 3 = 1$$

The solution to the system is \(x = 2\) and \(y = 1\), corresponding to the ordered pair \((x, y) = (2, 1)\).

Check in both original equations:

In (1):

$$1 = 2(2) - 3 = 4 - 3 = 1$$

In (2):

$$3(2) + 2(1) = 6 + 2 = 8$$

Both statements are true.

Example 12

Solve the following system of linear equations:

$$\begin{cases} 3(x - 1) + 2y = x + 1 \\ 2x - (y + 3) = 1 \end{cases}$$

Solution

Before applying the substitution method, clear parentheses and combine like terms to rewrite each equation in standard form \(Ax + By = C\).

Expand and simplify the first equation:

$$3x - 3 + 2y = x + 1$$

$$3x - x + 2y = 1 + 3$$

$$2x + 2y = 4 \quad (1')$$

Simplify equation (1') further by dividing all terms by 2:

$$x + y = 2 \quad (1'')$$

Expand and simplify the second equation, distributing the negative sign across the parentheses:

$$2x - y - 3 = 1$$

$$2x - y = 1 + 3$$

$$2x - y = 4 \quad (2')$$

With the simplified system in place, solve for y in equation (1''):

$$y = 2 - x \quad (3)$$

Substitute this expression into equation (2'):

$$2x - (2 - x) = 4$$

Distribute the negative sign and combine like terms:

$$2x - 2 + x = 4$$

$$3x - 2 = 4$$

Add 2 to both sides and solve for x:

$$3x = 4 + 2$$

$$3x = 6$$

$$x = \frac{6}{3} = 2$$

Substitute \(x = 2\) into equation (3) to compute y:

$$y = 2 - 2 = 0$$

The solution point is \((x, y) = (2, 0)\).

Check in the original equations:

In the first equation:

$$3(2 - 1) + 2(0) = 2 + 1 \implies 3 = 3$$

In the second equation:

$$2(2) - (0 + 3) = 4 - 3 = 1$$

Both statements hold true.

Example 13

$$\begin{cases} 15x - 11y = 32 \\ 7y - 9x = 8 \end{cases}$$

Solution

Assign a number to each equation:

$$15x - 11y = 32 \quad (1)$$

$$7y - 9x = 8 \quad (2)$$

Note that in equation (2) the variables appear in reverse order (\(y\) precedes \(x\)). Take care during substitution to place the expression into the matching variable.

Solve for x in equation (2), isolating the term with the negative coefficient:

$$7y - 8 = 9x \implies x = \frac{7y - 8}{9} \quad (3)$$

Substitute this expression in place of x in equation (1):

$$15\left(\frac{7y - 8}{9}\right) - 11y = 32$$

Simplify the factor 15 with the denominator 9 by dividing both by 3:

$$5\left(\frac{7y - 8}{3}\right) - 11y = 32$$

Distribute the factor of 5 in the numerator:

$$\frac{35y - 40}{3} - 11y = 32$$

Multiply the entire equation by 3 to eliminate the denominator:

$$(35y - 40) - 3(11y) = 3(32)$$

$$35y - 40 - 33y = 96$$

Combine like terms and add 40 to both sides:

$$2y - 40 = 96$$

$$2y = 96 + 40$$

$$2y = 136$$

$$y = \frac{136}{2} = 68$$

Substitute \(y = 68\) into equation (3) to calculate x:

$$x = \frac{7(68) - 8}{9} = \frac{476 - 8}{9} = \frac{468}{9} = 52$$

The solution to the system is \(x = 52\) and \(y = 68\), corresponding to the point \((x, y) = (52, 68)\).

Check the result in both original equations:

In (1):

$$15(52) - 11(68) = 780 - 748 = 32$$

In (2):

$$7(68) - 9(52) = 476 - 468 = 8$$

Both relationships are confirmed.

Example 14

Determine the values of m and n that satisfy the system:

$$\begin{cases} 2m + 5n = 14 \\ 5m + 2n = 23 \end{cases}$$

Solution

The substitution method works the same way regardless of the variable letters used. Identify the equations:

$$2m + 5n = 14 \quad (1)$$

$$5m + 2n = 23 \quad (2)$$

Solve for m in equation (1):

$$2m = 14 - 5n \implies m = \frac{14 - 5n}{2} \quad (3)$$

Substitute this expression into equation (2):

$$5\left(\frac{14 - 5n}{2}\right) + 2n = 23$$

Multiply the numerator by 5:

$$\frac{70 - 25n}{2} + 2n = 23$$

Multiply every term in the equation by 2 to clear fractions:

$$(70 - 25n) + 2(2n) = 2(23)$$

$$70 - 25n + 4n = 46$$

Combine like terms and group the constants:

$$70 - 21n = 46$$

$$-21n = 46 - 70$$

$$-21n = -24$$

$$n = \frac{-24}{-21} = \frac{8}{7}$$

Substitute \(n = \frac{8}{7}\) into equation (3) to solve for m:

$$m = \frac{14 - 5\left(\frac{8}{7}\right)}{2} = \frac{14 - \frac{40}{7}}{2}$$

Express 14 as an equivalent fraction with denominator 7 (\(14 = \frac{98}{7}\)):

$$m = \frac{\frac{98}{7} - \frac{40}{7}}{2} = \frac{\frac{58}{7}}{2} = \frac{58}{14} = \frac{29}{7}$$

The exact solution is \(m = \frac{29}{7}\) and \(n = \frac{8}{7}\), written as the ordered pair \((m, n) = \left(\frac{29}{7}, \frac{8}{7}\right)\).

Check in the original equations:

In (1):

$$2\left(\frac{29}{7}\right) + 5\left(\frac{8}{7}\right) = \frac{58}{7} + \frac{40}{7} = \frac{98}{7} = 14$$

In (2):

$$5\left(\frac{29}{7}\right) + 2\left(\frac{8}{7}\right) = \frac{145}{7} + \frac{16}{7} = \frac{161}{7} = 23$$

The system is verified.

Example 15

Solve the following system of three linear equations in three variables:

\(\begin{cases} x + y + z = 6 \\ 2x + y - z = 1 \\ x - y + z = 2 \end{cases}\)

Solution

While 3×3 systems are frequently solved using Gaussian elimination or matrices, the substitution method can be applied systematically: isolate one variable in a chosen equation and substitute it into the other two, reducing the problem to an equivalent 2×2 system.

Number the three equations:

$$x + y + z = 6 \quad (1)$$

$$2x + y - z = 1 \quad (2)$$

$$x - y + z = 2 \quad (3)$$

Solve for x in equation (1):

$$x = 6 - y - z \quad (4)$$

Substitute the expression for x into equations (2) and (3):

In equation (2):

$$2(6 - y - z) + y - z = 1$$

$$12 - 2y - 2z + y - z = 1$$

$$-y - 3z = 1 - 12$$

$$-y - 3z = -11 \implies y + 3z = 11 \quad (5)$$

In equation (3):

$$(6 - y - z) - y + z = 2$$

$$6 - 2y = 2$$

$$-2y = 2 - 6$$

$$-2y = -4 \implies y = \frac{-4}{-2} = 2$$

Here, the variable z canceled out directly in equation (3), immediately yielding \(y = 2\). Substitute this value into equation (5) to solve for z:

$$2 + 3z = 11$$

$$3z = 11 - 2$$

$$3z = 9$$

$$z = \frac{9}{3} = 3$$

With \(y = 2\) and \(z = 3\) determined, find the remaining variable by substituting into equation (4):

$$x = 6 - 2 - 3 = 1$$

The unique solution to the system is the ordered triple \((x, y, z) = (1, 2, 3)\).

Check the values in all three original equations:

In (1):

$$1 + 2 + 3 = 6$$

In (2):

$$2(1) + 2 - 3 = 2 + 2 - 3 = 1$$

In (3):

$$1 - 2 + 3 = 2$$

All three statements are satisfied simultaneously.

Example 16

The sum of two integers is 45. If one integer is 6 more than twice the other, what are the two numbers?

Solution

Define the variables for the problem:

• x: first integer.
• y: second integer.

Translate the given conditions into algebraic equations:

$$x + y = 45 \quad (1)$$

$$x = 2y + 6 \quad (2)$$

Equation (2) already has the variable x isolated. Substitute this expression directly into equation (1):

$$(2y + 6) + y = 45$$

Combine like terms and subtract 6 from both sides:

$$3y + 6 = 45$$

$$3y = 45 - 6$$

$$3y = 39$$

$$y = \frac{39}{3} = 13$$

Substitute \(y = 13\) into equation (2) to calculate x:

$$x = 2(13) + 6 = 26 + 6 = 32$$

The required numbers are 32 and 13.

Check the conditions: their sum is \(32 + 13 = 45\), and the larger number is 6 more than twice the smaller number: \(2(13) + 6 = 32\).

Example 17

The perimeter of a rectangular lot is 56 meters. If the length is 4 meters more than three times the width, find the dimensions of the lot.

Solution

Define variables for the geometric dimensions:

• L: length of the rectangle in meters.
• W: width of the rectangle in meters.

The perimeter of a rectangle is given by the formula \(P = 2L + 2W\). Set up the system using the given information:

$$2L + 2W = 56 \quad (1)$$

$$L = 3W + 4 \quad (2)$$

Simplify equation (1) by dividing all terms by 2:

$$L + W = 28 \quad (1')$$

Substitute the expression for L from equation (2) into equation (1'):

$$(3W + 4) + W = 28$$

Combine like terms and isolate W:

$$4W + 4 = 28$$

$$4W = 28 - 4$$

$$4W = 24$$

$$W = \frac{24}{4} = 6$$

Calculate the length L by substituting \(W = 6\) into equation (2):

$$L = 3(6) + 4 = 18 + 4 = 22$$

The dimensions of the lot are 22 meters in length and 6 meters in width.

Check with the original perimeter: \(2(22) + 2(6) = 44 + 12 = 56\text{ meters}\).

Example 18

A theater sold a total of 150 tickets for a performance, consisting of general admission tickets at $8 each and student tickets at $5 each. If total ticket sales were $960, how many tickets of each type were sold?

Solution

Define the number of each type of ticket:

• x: number of general admission tickets sold.
• y: number of student tickets sold.

Write one equation for the total number of tickets and another for total revenue:

$$x + y = 150 \quad (1)$$

$$8x + 5y = 960 \quad (2)$$

Solve for x in equation (1):

$$x = 150 - y \quad (3)$$

Substitute this expression into equation (2):

$$8(150 - y) + 5y = 960$$

Apply the distributive property:

$$1200 - 8y + 5y = 960$$

$$1200 - 3y = 960$$

Subtract 1200 from both sides and solve for y:

$$-3y = 960 - 1200$$

$$-3y = -240$$

$$y = \frac{-240}{-3} = 80$$

Calculate x by substituting \(y = 80\) into equation (3):

$$x = 150 - 80 = 70$$

The theater sold 70 general admission tickets and 80 student tickets.

Check the revenue using the unit prices: \(70(\$8) + 80(\$5) = \$560 + \$400 = \$960\).

Example 19

Currently, a father is three times as old as his daughter. In 12 years, the father will be twice as old as his daughter will be at that time. How old is each person today?

Solution

Define variables for their current ages:

• F: father's current age.
• D: daughter's current age.

In 12 years, the father's age will be \(F + 12\) and the daughter's age will be \(D + 12\). Set up the two equations:

$$F = 3D \quad (1)$$

$$F + 12 = 2(D + 12) \quad (2)$$

Substitute \(F = 3D\) from equation (1) into equation (2):

$$3D + 12 = 2(D + 12)$$

Apply the distributive property on the right side:

$$3D + 12 = 2D + 24$$

Group terms with D on the left and constant terms on the right:

$$3D - 2D = 24 - 12$$

$$D = 12$$

Calculate the father's age by substituting \(D = 12\) into equation (1):

$$F = 3(12) = 36$$

Currently, the father is 36 years old and the daughter is 12 years old.

Check the future condition: in 12 years, the father will be 48 and the daughter will be 24, satisfying \(48 = 2(24)\).

Example 20

A fitness center offers two monthly membership options. Plan A charges a basic registration fee of $20 plus $4 per fitness class attended. Plan B has no registration fee but charges a flat monthly fee of $35 plus $1.50 per class attended. How many classes must a member attend in a month for both plans to cost the exact same amount?

Solution

Define the variables for the cost model:

• C: total monthly cost in dollars.
• n: number of fitness classes attended.

Write the cost equation for each plan:

$$C = 20 + 4n \quad (1)$$

$$C = 35 + 1.5n \quad (2)$$

Since both equations express the value of the same dependent variable C, substitute the expression from equation (1) into equation (2):

$$20 + 4n = 35 + 1.5n$$

Group terms containing n on the left and constants on the right:

$$4n - 1.5n = 35 - 20$$

$$2.5n = 15$$

Divide by 2.5 to solve for n:

$$n = \frac{15}{2.5} = 6$$

Substitute \(n = 6\) into equation (1) to determine the corresponding total cost:

$$C = 20 + 4(6) = 20 + 24 = 44$$

A member must attend 6 classes per month for both plans to cost the same amount, which totals $44.

Check with Plan B for 6 classes: \(C = 35 + 1.5(6) = 35 + 9 = 44\text{ dollars}\).