Describing Solutions to Systems of Linear Equations

a

We want to find the number of solutions for the following system of equations. ${2 y = 4 x - 6 6 x = 3 y + 7 (I) (II)$ To determine how many solutions this system has, we will solve it by substitution. Doing so will result in one of three cases.

Result of solving by substitution	Number of solutions
A value for $x$ and $y$ is determined.	One solution
An identity is found, such as $2 = 2 .$	Infinitely many solutions
A contradiction is found, such as $2 \neq = 3 .$	No solution

Thus, we need to solve the system of equations before we can make our conclusion. When solving a system of equations using substitution, there are three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.

Let's isolate

y

in Equation (I).

2 y = 4 x - 6 \Leftrightarrow y = 2 x - 3

Let's substitute

2 x - 3

for

y

in Equation (II) and solve the resulting equation for

x .

6 x = 3 y + 7

Substitute

y = 2 x - 3

6 x = 3 (2 x - 3) + 7

DistrDistribute

3

6 x = 6 x - 9 + 7

AddTermsAdd terms

6 x = 6 x - 2

SubEqn

LHS - 6 x = RHS - 6 x

0 \neq = - 2

Solving this system resulted in a contradiction. This means that the system has no solution.

b

Here we have been asked to determine how many solutions the system ${x + \frac{2}{3} y = 3 - 6 x = 4 y + 5 (I) (II)$ has. We will first solve the system by substitution. After that we will use the following table to find the number of solutions.

Result of solving by substitution	Number of solutions
A value for $x$ and $y$ is determined.	One solution
An identity is found, such as $2 = 2 .$	Infinitely many solutions
A contradiction is found, such as $2 \neq = 3 .$	No solution

Let's begin the solution by isolating

x

in Equation (I).

x + \frac{2}{3} y = 3 \Leftrightarrow x = 3 - \frac{2}{3} y

Next step of the solution is to substitute

x

for

3 - \frac{2}{3} y

in Equation (II) and solve the resulting equation.

- 6 x = 4 y + 5

Substitute

x = 3 - \frac{2}{3} y

- 6 (3 - \frac{2}{3} y) = 4 y + 5

DistrDistribute

(- 6)

- 18 + 4 y = 4 y + 5

SubEqn

LHS - 4 y = RHS - 4 y

- 18 \neq = 5

Here we ended up with a contradiction,

- 18 \neq = 5 .

Thus, the system has no solution.

Describing Solutions to Systems of Linear Equations 1.4 - Solution