Question

determining the number of solutions of a system of linear equations
how many solutions will each system of linear equations have? match the systems with the correct number of solutions.
$y = -2x + 5$ and $2x + y = -7$
$y = x + 6$ and $3x - 3y = -18$
$y = -4x + 11$ and $-4x + y = 11$
no solution
one solution
infinitely many solutions

Explanation:

Step1: Analyze first system

First system: $y = -2x + 5$ and $2x + y = -7$. Substitute $y$ from first equation into second:
$2x + (-2x + 5) = -7$
$2x - 2x + 5 = -7$
$5 = -7$
This is a false statement, so no solution.

Step2: Analyze second system

Second system: $y = x + 6$ and $3x - 3y = -18$. Substitute $y$ into second equation:
$3x - 3(x + 6) = -18$
$3x - 3x - 18 = -18$
$-18 = -18$
This is always true, so infinitely many solutions.

Step3: Analyze third system

Third system: $y = -4x + 11$ and $-6x + y = 11$. Substitute $y$ into second equation:
$-6x + (-4x + 11) = 11$
$-10x + 11 = 11$
$-10x = 0$
$x = 0$
Substitute $x=0$ back to find $y = -4(0)+11=11$. There is one unique solution.

Answer:

• $y = -2x + 5$ and $2x + y = -7$ → no solution
• $y = x + 6$ and $3x - 3y = -18$ → infinitely many solutions
• $y = -4x + 11$ and $-6x + y = 11$ → one solution