I have done them all but 2 i just don't get and 2 I need to check if they are correct
1)
2x+3y=31
y=x+7
2)
y=3x
x=3y
2006-12-04 14:37:07 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Hey I'm not forcing you to answer and I'm sorry no need to be rude I'm really struggling with math it's not my best subject I'm really trying I had 20 problems to do and I'm only asking for 2 earlier I only asked for one to make sure it was correct
2006-12-04 14:44:22 · update #1
2x + 3y = 31
y = x + 7
First, put x and y terms on the left hand side.
2x + 3y = 31
x - y = -7
Now, multiply the second equation by 2.
2x + 3y = 31
2x - 2y = -14
Subtract the equations to eliminate one variable.
(2x + 3y) - (2x - 2y) = 31 - (-14)
5y = 45
y = 9
Now that you solved for y, you can solve for x using any one of the two equation. Let's use x - y = -7
x - 9 = -7
x = 2
Therefore, x = 9 and y = 9.
For #2, again, you have to put all x terms and y terms to the left hand side.
3x - y = 0
x - 3y = 0
Multiply the second equation by 3.
3x - y = 0
3x - 9y = 0
Now, subtract the two equations.
(3x - y) - (3x - 9y) = 0
-y + 9y = 0
8y = 0, so y = 0
You'll learn through substitution that x = 0 as well.
Therefore, x = 0 and y = 0
2006-12-04 14:42:03 · answer #1 · answered by Puggy 7 · 0⤊ 1⤋
1) y=9
x=2
2) y=0 and x=0
2006-12-04 14:40:56 · answer #2 · answered by Anonymous · 0⤊ 1⤋
enable the present age of the guy be x years. enable the present age of the son be y years. in accordance to first situation: (x-2) =3(y-2)^2 x-2=3(y^2-4y+4) x-2=3y^2-12y+12 x=3y^2-12y+14-------equation a million in accordance to the second one situation: x+3= 4(y+3) x+3=4y+12 x=4y+9----------equation 2 Subtracting equation 2 from a million,we will get 3y^2-16y+5=0----------equation 3. (3y-a million)(y-5)=0 y=a million/3 or 5 considering that, age can't be a million/3. So y=5. Substituting the cost of y in equation 2, we will get the cost of x as 29. hence, the age of pop and son is 29 and 5 years respectively. i wish that is sparkling adequate.
2016-10-16 11:49:23 · answer #3 · answered by ? 4 · 0⤊ 0⤋
2x+3y=31
y=x+7
plug y into the first equation.
2x+3(x+7)=31
2x+3x+21=31
5x=10
x=2
Plug x into the 2nd equation...
y=2+7=9
x=2 y=9
number 2 is x=0 y=0
2006-12-04 14:42:37 · answer #4 · answered by jennytkd13 3 · 0⤊ 1⤋
1 is x = 2
2 is y = x
2006-12-04 14:42:02 · answer #5 · answered by pixi_doll 3 · 0⤊ 1⤋
1) x = 2, y =9
2) x=y=0
2006-12-04 14:42:23 · answer #6 · answered by shamu 2 · 0⤊ 1⤋
2x+3y=31
3y=-2x+31 move variables so y is by itself
y=(2/3)x+(31/3) divide 3 from the y
is that what you are asking?
2006-12-04 14:42:14 · answer #7 · answered by Nerds Gone Wild 2 · 0⤊ 1⤋
Pay attention in class then you wouldnt have to cheat by asking on yahoo...and its not that hard by the way
2006-12-04 14:39:18 · answer #8 · answered by Joel H 4 · 0⤊ 2⤋
You and your math teacher need to have a LONG talk.
2006-12-04 14:38:49 · answer #9 · answered by Anonymous · 1⤊ 1⤋
1)
2x+3y=31
2x+14=2y
-------------
3y-14=31-2y
5y=45
y=9
x=2
2)
y=3x
3x=9y
y=9y
8y=0
y=0
x=0
2006-12-04 14:41:28 · answer #10 · answered by heartsensei 4 · 0⤊ 1⤋