Question

practice determining the number of solutions to a system of linear equations

1. ( y = x - 5 )

( x = y + 5 )
(circ) no solution (circ) one solution (circ) infinitely many solutions

1. ( y = x )

( y = 8x )
(circ) no solution (circ) one solution (circ) infinitely many solutions

1. ( y = x + 6 )

( y = x + 1 )
(circ) no solution (circ) one solution (circ) infinitely many solutions

1. ( y = \frac{1}{3}x + 2 )

( y = -\frac{1}{3}x + 2 )
(circ) no solution (circ) one solution (circ) infinitely many solutions

Explanation:

First System: \( y = x - 5 \) and \( x = y + 5 \)

Step1: Rewrite the second equation

Rewrite \( x = y + 5 \) as \( y = x - 5 \) (subtract 5 from both sides).

Step2: Compare with the first equation

The first equation is \( y = x - 5 \), so both equations are identical.

Step1: Substitute \( y \) from first into second

Substitute \( y = x \) into \( y = 8x \), we get \( x = 8x \).

Step2: Solve for \( x \)

Subtract \( x \) from both sides: \( 0 = 7x \), so \( x = 0 \). Then \( y = 0 \).

Step1: Subtract the two equations

Subtract \( y = x + 1 \) from \( y = x + 6 \): \( 0 = 5 \), which is false.

Step2: Determine the number of solutions

Since the equation is false, there are no solutions.

Answer:

Infinitely Many Solutions

Second System: \( y = x \) and \( y = 8x \)