To determine which system of equations has (5, -3) as a solution, let's evaluate each equation for the point (5, -3), where \( x = 5 \) and \( y = -3 \).
System 1
-
Equation: \( y = \frac{2}{3}x + 6 \)
\(-3 = \frac{2}{3}(5) + 6\)
\(-3 = \frac{10}{3} + 6\)
\(-3 \neq \frac{10}{3} + \frac{18}{3}\) (not satisfied) -
Equation: \( 3y + 6x - 18 = 0 \)
\( 3(-3) + 6(5) - 18 = 0 \)
\(-9 + 30 - 18 = 0\)
\(3 = 0\) (not satisfied)
System 2
-
Equation: \( y = x - 8 \)
\(-3 = 5 - 8\)
\(-3 = -3\) (satisfied) -
Equation: \( 2x + y = 7 \)
\( 2(5) + (-3) = 7\)
\( 10 - 3 = 7\)
\( 7 = 7\) (satisfied)
System 3
-
Equation: \( y = x + 2 \)
\(-3 = 5 + 2\)
\(-3 \neq 7\) (not satisfied) -
Equation: \( y = x + 5 \)
\(-3 = 5 + 5\)
\(-3 \neq 10\) (not satisfied)
System 4
-
Equation: \( y = -\frac{1}{2}x + 6 \)
\(-3 = -\frac{1}{2}(5) + 6\)
\(-3 = -\frac{5}{2} + 6\)
\(-3 \neq -\frac{5}{2} + \frac{12}{2}\) (not satisfied) -
Equation: \( y = 3x - 1 \)
\(-3 = 3(5) - 1\)
\(-3 \neq 15 - 1\) (not satisfied)
Conclusion
The system of equations that has the point (5, -3) as a solution is:
System 2:
- \( y = x - 8 \)
- \( 2x + y = 7 \)