Solve the systems : 1) x = 3y 3x + y = 10 2) 2x + 7y = 1 2x - 2y = 9
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x = 3y........(1)
3x + y = 10.........(2)
Using the substitution method, we will substitute x values from (1) in equation (2):
==> 3x + y = 10
==> 3(3y) + y = 10
==> 9x + y =...
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For the system of equations
x=3y and 3x+y=10
3x+y=10
=>9y+y=10
=>10y=10
=>y=1
=>x=y = 3
Therefore for the first system x=1 and y=3
For the second system of equations: 2x+7y=1 and 2x-2y=9
Subtract the 2nd from the first, we get 7y+2y=1-9
=>9y=-8
=>y=-8/9
Substitute this is 2x-2y=9
2x-2*(-8/9)=9
=>2x+16/9=9
=>x=(9-16/9)/2
=>x=(65/18)
Therefore for this system of equations x=65/18 and y=-8/9
1)
x = 3y
3x + y = 10
From the first equation, x = 3y. Substitute in the secon equation. We get: 3(3y)+3y = 10. So 9y+y = 10. Or 10y = 10, y =1. Put y = 1in the 1st equation: x = 3.
(x,y) = (3,1) is the solution.
2)
2x + 7y = 1
2x - 2y = 9
Subtract the 2nd equation from the 1st equation :
(2x+7y) - (2x-2y) = 1-9. X gets eliminated.
7y- -2y = 1-9 = -8
9y = -8, y = -8/9.
Put y = -8/9 and from the 2nd , 2x-2y = 9
2x-2(-8/9) = 9
2x= 9-16/9 = (81-16)/9
x = 65/18.
Therefore (x,y) = (65/18 , -8/9)
We'll solve the first system of equations using substitution technique:
x=3y
We'll substitute x by 3y, in the second equation of the system:
3*3y+y=10
We'll combine like terms:
9y+y=10
10y=10
We'll divide by 10 both sides:
y=1
Now, all we'll substitute the value for y into the first equation:
x=3y
x=3*1
x=3
The solution of the system is {(1 , 3)}.
Let's solve the second system of equations:
2x + 7y = 1
2x - 2y = 9
We'll solve this system using the substitution method:
2x=1-7y
Now, instead of 2x, we'll write in the second equation 1-7y.
1 - 7y - 2y =9
We'll combine like terms:
1 - 9y=9
We'll subtract 1 both sides:
-9y=8
We'll divide by -9 both sides:
y=-8/9
But 2x=1-7y and y=-8/9.
2x = 1 + 7*8/9
2x = 1 + 56/9
2x = (9+56)/9
2x = 65/9
We'll divide by 2:
x = 65/18
The solution of the system is: {(-8/9 , 65/18)}.
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