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
Values for $x$ and $y$ are determined.	One solution
An identity is found.	Infinitely many solutions
A contradiction is found.	No solution

This means that we should solve the given system of equations, then make our conclusion based on the result.

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.

Consider the given equations, we need to isolate one of the variables. Let's start by isolating

x

in Equation I and then substituting it into Equation II.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - x + \frac{1}{2} y = 13 x + 15 = \frac{1}{2} y (I) (II)

$(I):$ $LHS + x = RHS + x$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \frac{1}{2} y = 13 + x x + 15 = \frac{1}{2} y (I) (II)

$(I):$ $LHS - 13 = RHS - 13$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \frac{1}{2} y - 13 = x x + 15 = \frac{1}{2} y (I) (II)

$(I):$ Rearrange equation

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = \frac{1}{2} y - 13 x + 15 = \frac{1}{2} y (I) (II)

Now that we've isolated

x,

we can solve the system by substitution.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = \frac{1}{2} y - 13 x + 15 = \frac{1}{2} y (I) (II)

$(II):$ $x = \frac{1}{2} y - 13$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = \frac{1}{2} y - 13 \frac{1}{2} y - 13 + 15 = \frac{1}{2} y (I) (II)

$(II):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = \frac{1}{2} y - 13 \frac{1}{2} y + 2 = \frac{1}{2} y (I) (II)

$(II):$ $LHS - \frac{1}{2} y = RHS - \frac{1}{2} y$

⎩ ⎪ ⎨ ⎪ ⎧ x = \frac{1}{2} y - 13 2 \neq = 0 (I) (II)

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

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