Solve the system x+2y=3 4x+5y=6
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x+2y =3 ..........(1)
4x + 5y = 6........(2)
Let us use the substitution method:
from (1):
x = 3- 2y
Now substitute in (2):
4x + 5y = 6
4(3-2y) + 5y = 6
12 - 8y + 5y = 6
-3y = -6
y= 2
x = 3-2y = 3- 2*2 = -1
x= -1
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x+2y=3
4x+5y=6
multiply (x+2y=3) by 4 and subtract the equation 4x+5y=6
4x+8y=12
- 4x+5y=6
0x+3y=6
y=2
so plug in y in any equation to get x
x+4=3
x=-1
there you go :)
x + 2y = 3
4x + 5y = 6
First multiply every single digit in the first equation by 4
By doing that, your equation should look like
4x + 8y = 12
4x + 5y = 6 now, subtract 4x with 4x. Which means also subtract 8y with 5y and 12 with 6
By subtracting, your equation should look like
3y = 6 now divide both sides by 3
By dividing, your equation should look like
y = 2 which is your answer for " y "
Now, plug 2 into one of the equation
By doing that, you should get
x + 4 = 3 subtract both sides by 4
By subtracting, your equation should look like
x = -1 which is your answer for " x "
So your answer to this solution is x = -1 ; y = 2
x+2y=3
4x+5y=6
ELIMINATION:
(-4)(x+2y=3)
4x+5y=6
-4x-8y=-12
4x+5y=6 <---add both linear equations
0x-3y=-6
-3y=-6
y=2
Substitution:
4x+5y=6
y=2
4x+5(2)=6
4x+10=6
4x=-4
x=-1
SOLUTION SET: (-1,2)
x + 2y = 3 ... (1)
4x + 5y = 6 ... (2)
Multiplying equation (1) by 4:
4x + 8y = 12 ... (3)
Subtracting equation (2) from equation (3):
4x - 4x + 8y - 5y = 12 - 6
==> 3y = 6
Therefore:
y = 6/3 = 2
Substituting this value of y in equation (1):
x + 2*2 = 3
==> x = 3 - 4 = -1
Answer:
x = -1
y = 2
x+2y =3
4x+5y = 6.
Solution:
a1x+b1y =c1........(1)
a2x+b2y =c2.........(2)
Eliminate y by (1)*b2 - (2)*b1
( a1b2-a2b1)x= b2c1- b1c2. Now solve for x:
x = ( b2c1-b1c2)/(a1b2-a2b1)
Similarly eliminate x by the operation (1)*a2- (2)*a1
(a2b1-a1b2)y = a2c1-a1c2
y = (a2c1-a1c2)/(a2b1-a1b2)
In this case a1 = 1 , b1 =2 , c1 = 3, a2 = 4, b2 =5 , c2 =6.
So x= ( b2c1-b1c2)/(a1b2-a2b1) = (5*3-2*6)/(1*5 - 4*2) = 3/-3 = -1.
y = (a2c1-a1c2)(a2b1-a1b2) = (4*3-1*6)/(4*2-1*5) = 6/3 =2.
Therefore (x , y) = (-1 ,2)
We are given the equations
x+2y=3 ...(1)
and 4x+5y=6 ...(2)
Multiply (1) by 4 and subtract (2), we get
4x+8y-4x-5y=12-6
=>3y=6
=>y=2
Substituting y=2 in (1), x+4=3 or x=-1
We'll solve the system using the substitution method.
We'll note the equations of the system as:
x+2y=3 (1)
4x+5y=6 (2)
We'll re-write (1):
x = -2y+3 (3)
We'll substitute x in (2):
4(-2y+3)+5y=6
We'll remove the brackets:
-8y + 12 + 5y - 6 = 0
We'll combine like terms:
-3y + 6 = 0
We'll subtract 6 both sides:
-3y = -6
We'll divide by -3:
y = 2
We'll substitute y in (3):
x = -2y+3
x = -2*2 + 3
x = -4+3
x = -1
The solution of the system is: {(-1 , 2)}.
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