Respuesta :
Solution:
Given the system of equations:
[tex]\begin{gathered} 5x-y=-4\text{ -- equation 1} \\ x+6y=2\text{ --- equation 2} \end{gathered}[/tex]To determine if the ordered pair (4, 0) is a solution to the system of equations, we solve for x and y.
Thus, by the substitution method of solving simultaneous equations, we have
from equation 1,
[tex]\begin{gathered} make\text{ y the subject of formula,} \\ y=5x+4---\text{ equation 3} \end{gathered}[/tex]Substitute equation 3 into equation 2.
Thus, we have
[tex]\begin{gathered} x+6(5x+4)=2 \\ open\text{ parentheses,} \\ x+30x+24=2 \\ collect\text{ like terms,} \\ 31x=-22 \\ divide\text{ both sides by the coefficnet of x, which is 31} \\ \frac{31x}{31}=-\frac{22}{31} \\ \Rightarrow x=-\frac{22}{31} \end{gathered}[/tex]To solve for y, substitute the obtained value of x into equation 3.
Thus,
[tex]y=5(-\frac{22}{31})+4=\frac{14}{31}[/tex]Thus, the solution to the system of equations is
[tex](x,y)\Rightarrow(-\frac{22}{31},\frac{14}{31})[/tex]Hence, the ordered pair (4, 0) is NOT a solution to the system of equations.
