Answer :
We can use substitution.
We can substitute 2x + 3 for 'y' in the 2nd equation.
y = 3x + 5
2x + 3 = 3x + 5
Subtract 3 to both sides:
2x = 3x + 2
Subtract 3x to both sides:
-x = 2
Multiply -1 to both sides:
x = -2
Now we can plug this into any of the two equations to get our y-value:
y = 2x + 3
y = 2(-2) + 3
y = -4 + 3
y = -1
So our solution is (-2, -1)
We can substitute 2x + 3 for 'y' in the 2nd equation.
y = 3x + 5
2x + 3 = 3x + 5
Subtract 3 to both sides:
2x = 3x + 2
Subtract 3x to both sides:
-x = 2
Multiply -1 to both sides:
x = -2
Now we can plug this into any of the two equations to get our y-value:
y = 2x + 3
y = 2(-2) + 3
y = -4 + 3
y = -1
So our solution is (-2, -1)
[tex]y=2x+3\\
y=3x+5\\\\
2x+3=3x+5\\
2x-3x=5-3\\
-x=2\\
x=-2\\\\
y=2\cdot(-2)+3\\
y=-4+3\\
y=-1[/tex]