y = x + 2
y = 3x
Substitute 3x for y in the first equation.
3x = x + 2
sbutract x
2x = 2
x = 1
plug back into original:
y = 3x
y = 3(1)
y = 3
so the sol'n is (1,3)
2006-11-27 09:51:51
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answer #1
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answered by mattomynameo 4
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1) y = x + 2
2) y = 3x
Both equations are already solved for y in terms of x, so you can use either one to make a substitution into the other
)1 into )2
x + 2 = 3x
2 = 2x
1 = x
y = x + 2
y = 1 + 2
y = 3
Point of intersection is (1, 3)
2006-11-27 17:54:32
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answer #2
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answered by kindricko 7
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After substituting you have
3x = x + 2
subtract x from both sides
2x = 2
divide by 2
x = 1
Now go back and plug 1 in for x and solve for y.
y = 1 + 2
y = 3
The answer is (1, 3).
2006-11-27 17:52:32
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answer #3
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answered by Anonymous
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You have both already specified in terms of y, so just equate them:
3x = x + 2
Now solve by subtracting x from both sides:
2x = 2
Divide both sides by 2:
x = 1
Now plug this back into one of the equations to get y:
y = 3x
y = 3(1)
y = 3
So your answers are:
x = 1, y = 3
This is answer a) 1, 3
2006-11-27 17:53:43
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answer #4
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answered by Puzzling 7
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x=1
y=3
because if u put it as: 3x=x+2
then u just substituted y=3x into y=x+2
after that u just subtract x from the 3x to get 2x then u divide 2x into 2 to get 1
then u go back to y=3x and plug in 1 for x then multiply 3 to 1 and that's Ur answer
2006-11-27 17:57:31
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answer #5
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answered by tnjalj 1
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by substitution
plug in y=3x into first equation
3x = x + 2
2x = 2
x=1
y= 3*1 = 3
thus (1,3)
check
y=x+2
3 = 1+2
true
CHECK
2006-11-27 17:53:58
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answer #6
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answered by af12af3af 2
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If y = x + 2 and y = 3x, then we can substitute 3x for y in the first equation.
3x = x + 2
2x = 2
x = 1
If y = 3x, then y = 3.
The answer is (1,3)
2006-11-27 17:52:00
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answer #7
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answered by hokiejthweatt 3
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3=1+2 correct
0=0+2
1=3+2
1=1/3+2
3=1x x=3
0=0x x=1,2,3...
1=3x x=1/3
1=1/3x or 1=.333...x x=1/.333...
2006-11-27 18:04:47
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answer #8
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answered by Anonymous
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it is 1,3 by using substitution
2006-11-27 17:51:05
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answer #9
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answered by waitin_for_an_angel 1
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1,3
2006-11-27 17:51:22
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answer #10
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answered by Anonymous
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