Answer:
[tex]x=2[/tex], [tex]y=0[/tex], or as an ordered pair [tex](x,y)=(2,0)[/tex].
Step-by-step explanation:
We have the system of equations
[tex]5x + 2y = 10[/tex] equation (1)
[tex]3x + 2y = 6[/tex] equation (2)
Since both equations have the term [tex]2y[/tex], we are using the elimination method:
Step 1. Multiply equation (2) by -1 and add the result equation to equation (1):
[tex]\left \{ {{-1(3x + 2y = 6)} \atop +{5x + 2y = 10}} \right.[/tex]
[tex]\left \{ {{-3x -2y = -6)} \atop +{5x + 2y = 10}} \right.[/tex]
Now we can get rid of [tex]-2y[/tex]
[tex]2x=4[/tex]
[tex]x=\frac{4}{2}[/tex]
[tex]x=2[/tex]
Step 2. Replace the value [tex]x=2[/tex] in equation (1) to find the value of [tex]y[/tex]:
[tex]5x + 2y = 10[/tex]
[tex]5(2) + 2y = 10[/tex]
[tex]10 + 2y = 10[/tex]
[tex]2y = 0[/tex]
[tex]y = 0[/tex]
We can conclude that the solution of the system of equations is [tex]x=2[/tex], [tex]y=0[/tex], or as an ordered pair [tex](x,y)=(2,0)[/tex].
