bobpursey, In this problem you helped me with earlier:
Solve using any appropriate method.
y = 2× - 5,
y = ½× + 1
Substitution, substitute for y.
2×-5=½ × + 1
3/2 x=6
x=4
y= 2*4-5=3
How did you choose which method you were going to use between substitution and elimination? Could this have done with the elimination method? Thanks
Since both equations were in the form y = ...
he simply replaced the y value of the first equation with the y value of the second.
Some texts call this the "comparison" method
e.g.
if a = b
and c = b , then a = c
yes, you could use elimination after some doctoring of the equations
y = 2x - 5 ---> 2x - y = 5
y = (1/2)x + 1 ---> 2y = x + 2 --> x - 2y = -2
new first by 2: --> 4x - 2y = 10
new 2nd : ------> x - 2y = -2
subtract them:
3x = 12
x = 4
back in the original first:
y = 2(4) - 5 = 3
x=4 , y = 3
When solving a system of equations, you have multiple methods at your disposal, including substitution and elimination.
To determine the most appropriate method, you need to consider the given equations and the variables involved.
In this case, we have two equations:
y = 2x - 5
y = 1/2x + 1
Both variables, x and y, are already isolated on one side in each equation. No coefficient is equal in magnitude and opposite in sign for either x or y.
Given these conditions, both substitution and elimination methods are applicable.
For substitution method:
- You can solve one equation for one variable, and then substitute that expression into the other equation.
- This method works well when one equation is already solved for either x or y, like the second equation here.
For elimination method:
- You manipulate the equations to make one variable cancel out when you add or subtract the equations.
- This method works well when the coefficients of one variable in both equations are equal in magnitude and opposite in sign.
In this case, either method could have been used. The given equations are already in a suitable form for both methods, and the coefficients of x and y are different in each equation.
To demonstrate elimination method:
Given equations:
y = 2x - 5
y = 1/2x + 1
To use the elimination method, we can multiply the second equation by 2 to make the coefficients of x in both equations the same:
2(y) = 2(1/2x + 1)
2y = x + 2
Now, subtract the first equation from this modified second equation:
2y - (2x - 5) = x + 2 - (2x - 5)
This simplifies to:
2y - 2x + 5 = x + 2 - 2x + 5
Combine terms:
2y - 2x = x + 2 - 2x
Now, arrange the terms to isolate y:
2y = 3x - 2
Divide through by 2:
y = (3/2)x - 1
As you can see, we get the same equation for y as before.
At this point, you can solve for x by setting the two expressions for y equal to each other:
2x - 5 = 1/2x + 1
Solve for x:
2x - 1/2x = 1 + 5
(4/2)x - (1/2)x = 6
(7/2)x = 6
Divide by (7/2):
x = (6) / (7/2)
Simplify the division with the reciprocal of 7/2:
x = 12/7
Thus, using the elimination method, we find that x = 12/7.
To find y, substitute this value back into either equation, let's use the first equation:
y = 2(12/7) - 5
Simplify:
y = 24/7 - 35/7
y = -11/7
Thus, using the elimination method, we find that y = -11/7.
In conclusion, you can choose either the substitution or elimination method based on the initial conditions of the given equations. In this case, both methods were applicable, and either method would have led to the same solution.