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2.6.3 Finding Solutions to Unordered Sets of Equivalent Systems

Algebra 12.6.3 Finding Solutions to Unordered Sets of Equivalent Systems

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Activity

Your teacher will give you some slips of paper with systems of equations written on them. Each system represents a step in solving this system:

${\begin{matrix} \frac{4}{5} x + 6 y = 15 \\ - x + 18 y = 11 \end{matrix}$

Arrange the slips in the order that would lead to a solution. Be prepared to:

1.

Describe what move takes one system to the next system.

Compare your answer:

${\begin{matrix} \frac{4}{5} x + 6 y = 15 \\ - x + 18 y = 11 \end{matrix}$

Step 1 -

${\begin{matrix} 4 x + 30 y = 75 - x + 18 y = 11 \end{matrix}$

Step 2 -

${\begin{matrix} 4 x + 30 y = 75 - 4 x + 72 y = 44 \end{matrix}$

Step 3 -

${\begin{matrix} 4 x + 30 y = 75 102 y = 119 \end{matrix}$

Step 4 -

${\begin{matrix} 4 x + 30 y = 75 y = \frac{7}{6} \end{matrix}$

Step 5 -

${\begin{matrix} 4 x + 35 = 75 y = \frac{7}{6} \end{matrix}$

Step 6 -

${\begin{matrix} 4 x = 40 y = \frac{7}{6} \end{matrix}$

Step 7 -

${\begin{matrix} x = 10 y = \frac{7}{6} \end{matrix}$

2.

Explain why the systems are equivalent between the original system and step 1.

Compare your answer:

Your answer may vary, but here is a sample.

Multiply the first equation by 5.

3.

Explain why the systems are equivalent between the system in step 1 to the system in step 2.

Compare your answer:

Your answer may vary, but here is a samples.

Multiply the second equation by 4.

4.

Explain why the systems are equivalent between the system in step 2 to the system in step 3.

Compare your answer:

Your answer may vary, but here is a samples.

Add the two equations in Step 2.

5.

Explain why the systems are equivalent between the system in step 3 to the system in step 4.

Compare your answer:

Your answer may vary, but here is a samples.

Divide each side of $102 y = 119$ by 102 to solve for $y$ .

6.

Explain why the systems are equivalent between the system in step 4 to the system in step 5.

Compare your answer:

Your answer may vary, but here is a samples.

Substitute $\frac{7}{6}$ for y in the first equation and evaluating the expression.

7.

Explain why the systems are equivalent between the system in step 5 to the system in step 6.

Compare your answer:

Your answer may vary, but here is a samples.

From Step 5 to Step 6 - Subtract 35 from each side of the first equation.

8.

Explain why the systems are equivalent between the system in step 6 to the system in step 7.

Compare your answer:

Your answer may vary, but here is a samples.

From Step 6 to Step 7 - Divide each side of $4 x = 40$ by 4 to solve for $x$ .

Are you ready for more?

Extending Your Thinking

This system of equations has the solution $(5, - 2)$ : ${\begin{matrix} A x - B y = 24 \\ B x + A y = 31 \end{matrix}$

1.

Find the missing value of $A$ .

Compare your answer:

$A = 2$

2.

Find the missing value of $B$ .

Compare your answer:

$B = 7$

Self Check

Solve the system of equations using elimination.

$\left\{\begin{array}{l}7x\;–\;2y\;=\;3\\4x\;+\;5y\;=\;3.25\end{array}\right.$

$\left(\frac12,\;\frac14\right)$
$(0, -1.5)$
$(1, 2)$
$(7, -5)$

Additional Resources

Solving a System of Equations by Elimination Involving Fractions

You just ordered the steps for solving a system of equations using elimination. Try using these steps when the system has equations containing fractions.

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the least common denominator (LCD) of all the fractions in the equation.

Solve the system by elimination: ${\begin{matrix} x + \frac{1}{2} y = 6 \\ \frac{3}{2} x + \frac{2}{3} y = \frac{17}{2} \end{matrix}$

Step 1 - Check to see if each equation is in standard form $A x + B y = C$ . Both equations are in standard form.

$x + \frac{1}{2} y = 6$

$\frac{3}{2} x + \frac{2}{3} y = \frac{17}{2}$

Step 2 - To clear the fractions, multiply each equation by its LCD (Least Common Denominator).

$2 (x + \frac{1}{2} y = 6)$

$6 (\frac{3}{2} x + \frac{2}{3} y = \frac{17}{2})$

Step 3 - Simplify the equations.

$2 x + y = 12$

$9 x + 4 y = 51$

Step 4 - Eliminate one of the variables and simplify the equation.

$- 4 (2 x + y = 12) = - 8 x - 4 y = - 48$

Step 5 - Add the two equations to eliminate $y$ .

$\begin{matrix} - 8 x - 4 y & = - 48 \\ 9 x + 4 y & = 51 \\ x & = 3 \end{matrix}$

Step 6 - Solve for $y$ by substituting $x$ back into one of the equations.

$\begin{matrix} 2 x + y & = 12 \\ 2 (3) + y & = 12 \\ 6 + y & = 12 \\ y & = 6 \end{matrix}$

Step 7 - Write the solution as an ordered pair.

The ordered pair is $(3, 6)$ .

Step 8 - Check that the ordered pair is a solution to both original equations.

$\begin{matrix} \frac{3}{2} x + \frac{2}{3} y = \frac{17}{2} \\ x + \frac{1}{2} y \overset{?}{=} 6 & \frac{3}{2} (3) + \frac{2}{3} (6) \overset{?}{=} \frac{17}{2} \\ 3 + \frac{1}{2} (6) \overset{?}{=} 6 & \frac{9}{2} + 4 \overset{?}{=} \frac{17}{2} \\ 3 + 3 \overset{?}{=} 6 & \frac{9}{2} + \frac{8}{2} \overset{?}{=} \frac{17}{2} \\ 6 = 6 ✓ & \frac{17}{2} = \frac{17}{2} ✓ \end{matrix}$

The solution is $(3, 6)$ .

Try it

Try It: Solving a System of Equations by Elimination Involving Fractions

Solve the following system of equations.

${\begin{matrix} \frac{1}{3} x - \frac{1}{2} y = 1 \\ \frac{3}{4} x - y = \frac{5}{2} \end{matrix}$

Compare your answer:

Step 1 - Check to see if each equation is in standard form $A x + B y = C$ . Both equations are in standard form.

$\frac{1}{3} x - \frac{1}{2} y = 1 \frac{3}{4} x - y = \frac{5}{2}$

Step 2 - To clear the fractions, multiply each equation by its LCD (Least Common Denominator).

$6 (\frac{1}{3} x - \frac{1}{2} y) = 6 (1) 4 (\frac{3}{4} x - y) = 4 (\frac{5}{2})$

Step 3 - Simplify the equations.

$2 x - 3 y = 6$

$3 x - 4 y = 10$

Step 4 - To get equivalent coefficients of $x$ , we will need to multiply the first equation by 3 and the second equation by 2.

$3 (2 x - 3 y) = 3 (6)$

$2 (3 x - 4 y) = 2 (10)$

Step 5 - Simplify.

$6 x - 9 y = 18$

$6 x - 8 y = 20$

Step 6 - Subtract the equations to eliminate one variable.

$\begin{matrix} 6 x - 9 y & = 18 \\ 6 x - 8 y & = 20 \\ - y & = - 2 \\ y & = 2 \end{matrix}$

Step 7 - Solve for $x$ by substituting $y$ back into one of the equations.

$\begin{matrix} 2 x - 3 y & = 6 \\ 2 x - 3 (2) & = 6 \\ 2 x - 6 & = 6 \\ 2 x & = 12 \\ x & = 6 \end{matrix}$

Step 8 - Write the solution as an ordered pair.

The ordered pair is $(6, 2)$ .

Step 9 - Check that the ordered pair is a solution to both original equations.

$\begin{matrix} \frac{3}{4} - y = \frac{5}{2} \\ \frac{3}{4} (6) - 2 = \frac{5}{2} \\ \frac{1}{3} x - \frac{1}{2} \overset{?}{=} 1 & \frac{18}{4} - 2 \overset{?}{=} \frac{5}{2} \\ \frac{1}{3} (6) \frac{1}{2} (6) \overset{?}{=} 6 & \frac{18}{4} - \frac{8}{4} \overset{?}{=} \frac{5}{2} \\ 2 - 1 \overset{?}{=} 1 & \frac{9}{2} + \frac{10}{4} \overset{?}{=} \frac{5}{2} \\ 1 = 1 ✓ & \frac{5}{2} = \frac{5}{2} ✓ \end{matrix}$

The solution is $(6, 2)$ .

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