Question

question 1-1
for each system of equations, indicate whether it has infinitely many solutions, one solution, or no solutions.
match each system of equations with its number of solutions.
$x + 3y = -15$ $2x + 6y = -30$
$y = -2x + 5$ $y = -2x - 8$
$2x - 3y = -4$ $3x - y = 1$
infinitely many solutions
one solution
no solutions

Explanation:

System of Equations 1: \( x + 3y = -15 \) and \( 2x + 6y = -30 \)

Step 1: Simplify the second equation

Divide the second equation \( 2x + 6y = -30 \) by 2: \( x + 3y = -15 \)

Step 2: Compare with the first equation

The first equation is \( x + 3y = -15 \), so both equations are identical.

Step 1: Analyze the slopes and y - intercepts

The slope - intercept form of a line is \( y=mx + b \), where \( m \) is the slope and \( b \) is the y - intercept. For \( y=-2x + 5 \), \( m=-2 \) and \( b = 5 \). For \( y=-2x-8 \), \( m=-2 \) and \( b=-8 \).

Step 2: Determine the number of solutions

Since the slopes are equal (\( m=-2 \)) and the y - intercepts are different (\( 5
eq - 8\)), the two lines are parallel and do not intersect.

Step 1: Solve the second equation for \( y \)

From \( 3x - y=1 \), we get \( y=3x - 1 \)

Step 2: Substitute \( y = 3x-1 \) into the first equation

Substitute into \( 2x-3y=-4 \): \( 2x-3(3x - 1)=-4 \)
Expand: \( 2x-9x + 3=-4 \)
Combine like terms: \( - 7x+3=-4 \)
Subtract 3 from both sides: \( -7x=-4 - 3=-7 \)
Divide both sides by - 7: \( x = 1 \)

Step 3: Find the value of \( y \)

Substitute \( x = 1 \) into \( y=3x - 1 \), we get \( y=3(1)-1=2 \)
Since we have a unique pair \( (x = 1,y = 2) \) that satisfies both equations.

Answer:

infinitely many solutions

System of Equations 2: \( y = -2x + 5 \) and \( y = -2x - 8 \)