Exercise 22 Page 382

To determine how many solutions this system has, we will solve it using the Elimination Method. 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≠ 3.	No solution

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

Solve by Elimination

To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. 3x+ 4y=24 & (I) 6x+ 8y=24 & (II) In its current state, this will not happen. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply (I) by -2 both the x-terms and the y-terms will have opposite coefficients. - 2(3 x+4 y)=- 2(24) 6 x+8 y=24 ⇒ - 6x- 8y=- 48 6x+ 8y=24 With this, we can see that both the x- and y-terms will eliminate each other if we add (I) to (II).

- 6x-8y=- 48 6x+8y=24

SysEqnAdd

(II):Add (I)

- 6x-8y=- 48 6x+8y+( - 6x-8y)=24+( - 48)

RemovePar

(II): Remove parentheses

- 6x-8y=- 48 6x+8y-6x-8y=24-48

AddSubTerms

(II): Add and subtract terms

- 6x-8y=- 48 0≠ - 24

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