Substitution? I would solve for x in the first equation:
x=6-2y check that. Then put that for x into the second equation...
4(6-2y)+16y=8 and solve for y.
-3x-y=3 3x+6y=48
-3x-5y=-21 -5x+6y=32
2x+4y=12
4x+16y=8
x=6-2y check that. Then put that for x into the second equation...
4(6-2y)+16y=8 and solve for y.
1. Solve one equation for one variable
Choose one of the equations and isolate one of the variables in terms of the other variable.
For the first system of equations:
-3x - y = 3 ... Equation 1
3x + 6y = 48 ... Equation 2
From Equation 1, isolate y:
y = -3x + 3
For the second system of equations:
-3x - 5y = -21 ... Equation 3
-5x + 6y = 32 ... Equation 4
From Equation 3, isolate y:
y = (-3x + 21)/5
For the third system of equations:
2x + 4y = 12 ... Equation 5
4x + 16y = 8 ... Equation 6
From Equation 5, isolate y:
y = (12 - 2x)/4
2. Substitute
Take the expression for the isolated variable from Step 1 and substitute it into the other equation.
For the first system:
Substitute y = -3x + 3 into Equation 2:
3x + 6(-3x + 3) = 48
Simplify and solve for x:
3x - 18x + 18 = 48
-15x = 48 - 18
-15x = 30
x = 30/(-15)
x = -2
Now substitute the value of x back into one of the original equations, Equation 1, to solve for y:
-3(-2) - y = 3
6 - y = 3
-y = 3 - 6
-y = -3
y = (-3)/(-1)
y = 3
So the solution to the first system of equations is x = -2 and y = 3.
For the second system:
Substitute y = (-3x + 21)/5 into Equation 4:
-5x + 6((-3x + 21)/5) = 32
Simplify and solve for x:
-5x - 18x + 126 = 160
-23x = 160 - 126
-23x = 34
x = 34/(-23)
x = -34/23
Now substitute the value of x back into one of the original equations, Equation 3, to solve for y:
-3((-34)/23) - 5y = -21
(-102)/23 - 5y = -21
-102 - 115y = -483
-115y = -483 + 102
-115y = -381
y = (-381)/(-115)
y = 381/115
So the solution to the second system of equations is x = -34/23 and y = 381/115.
For the third system:
Substitute y = (12 - 2x)/4 into Equation 6:
4x + 16((12 - 2x)/4) = 8
Simplify and solve for x:
4x + 48 - 8x = 8
-4x = 8 - 48
-4x = -40
x = (-40)/(-4)
x = 10
Now substitute the value of x back into one of the original equations, Equation 5, to solve for y:
2(10) + 4y = 12
20 + 4y = 12
4y = 12 - 20
4y = -8
y = (-8)/4
y = -2
So the solution to the third system of equations is x = 10 and y = -2.
Therefore, the solutions for the three given systems of equations using the substitution method are:
System 1: x = -2 and y = 3
System 2: x = -34/23 and y = 381/115
System 3: x = 10 and y = -2