Solve the two linear simultaneous equation, y + 2x -1=0 and 2y - 3x + 5=0
To solve the simultaneous equations:
1) y + 2x - 1 = 0
2) 2y - 3x + 5 = 0
We can use the method of substitution or elimination.
Method 1: Substitution
From equation 1, let's solve for y in terms of x:
y = -2x + 1
Now, substitute this value of y into equation 2:
2(-2x + 1) - 3x + 5 = 0
Simplify the equation:
-4x + 2 - 3x + 5 = 0
-7x + 7 = 0
-7x = -7
x = 1
Substitute this value of x back into equation 1 to solve for y:
y = -2(1) + 1
y = -2 + 1
y = -1
Therefore, the solution to the simultaneous equations is x = 1 and y = -1.
Method 2: Elimination
Multiply equation 1 by 2 and equation 2 by 3 to make the coefficients of x in both equations equal:
2(y + 2x - 1) = 0
3(2y - 3x + 5) = 0
Simplify the equations:
2y + 4x - 2 = 0
6y - 9x + 15 = 0
Now, we can combine these equations by adding them together:
(2y + 6y) + (4x - 9x) - (2 + 15) = 0
8y - 5x - 17 = 0
Rearrange this equation to isolate y:
8y = 5x + 17
y = (5/8)x + 17/8
So, we have derived the equation of a straight line that describes y in terms of x. This line has the same solution as the original simultaneous equations.
Therefore, there are infinitely many solutions to this system of equations, represented by the equation y = (5/8)x + 17/8.
To solve the system of linear equations, we can use the method of substitution or elimination. I will demonstrate the elimination method.
Step 1: Rearrange the equations to make the coefficients of either x or y the same for both equations:
Equation 1: y + 2x - 1 = 0
Equation 2: 2y - 3x + 5 = 0
Let's eliminate the y-term.
Step 2: Multiply both sides of Equation 1 by 2 to make the y-terms cancel out:
2(y + 2x - 1) = 2(0)
2y + 4x - 2 = 0
4x + 2y - 2 = 0 (Equation 3)
Step 3: Subtract Equation 3 from Equation 2 to eliminate the y-term:
(2y - 3x + 5) - (4x + 2y - 2) = 0
2y - 3x + 5 - 4x - 2y + 2 = 0
-3x - 4x + 2y + 2y + 5 + 2 = 0
-7x + 4y + 7 = 0 (Equation 4)
Step 4: Now we have two linear equations:
Equation 3: 4x + 2y - 2 = 0
Equation 4: -7x + 4y + 7 = 0
Step 5: Solve the system of linear equations using any method of your preference (e.g., substitution, elimination, or matrix):
Let's solve using the elimination method.
Multiply Equation 3 by 7 and Equation 4 by 4 to make the coefficients of x or y the same (opposite in sign):
28x + 14y - 14 = 0 (Equation 5)
-28x + 16y + 28 = 0 (Equation 6)
Adding Equation 5 and Equation 6, we can eliminate the x-term:
(28x + 14y - 14) + (-28x + 16y + 28) = 0
28x - 28x + 14y + 16y - 14 + 28 = 0
30y + 14 = 0
30y = -14
y = -14/30
y = -7/15
Step 6: Substitute the value of y into Equation 3 or 4 to find x:
Let's substitute it into Equation 3:
4x + 2(-7/15) - 2 = 0
4x - 14/15 - 2 = 0
4x - 14/15 = 2
4x = 2 + 14/15
4x = 30/15 + 14/15
4x = 44/15
x = 44/(4*15)
x = 11/15
Therefore, the solution to the system of linear equations is:
x = 11/15
y = -7/15
To solve the system of linear equations:
Equation 1: y + 2x - 1 = 0
Equation 2: 2y - 3x + 5 = 0
We can use the method of substitution or the method of elimination. Let's use the method of elimination for this example.
Step 1: Multiply Equation 1 by 3 to eliminate the x term in Equation 2.
Equation 1: 3(y + 2x - 1) = 0
=> 3y + 6x - 3 = 0
Equation 2: 2y - 3x + 5 = 0
Step 2: Add Equation 1 to Equation 2 to eliminate the x term.
(3y + 6x - 3) + (2y - 3x + 5) = 0
Simplifying the equation gives:
3y + 2y + 6x - 3x - 3 + 5 = 0
Combine like terms:
5y + 3x + 2 = 0
Step 3: Rearrange the equation obtained to solve for y.
5y = -3x - 2
Divide both sides by 5:
y = (-3x - 2) / 5
Step 4: Substitute the value of y in Equation 1 to solve for x.
y + 2x - 1 = 0
Substitute (-3x - 2) / 5 for y:
((-3x - 2) / 5) + 2x - 1 = 0
Step 5: Simplify and solve for x.
Multiply through by 5 to get rid of the fraction:
-3x - 2 + 10x - 5 = 0
Combine like terms:
7x - 7 = 0
Add 7 to both sides:
7x = 7
Divide by 7:
x = 1
Step 6: Substitute the value of x in Equation 1 or Equation 2 to solve for y.
Using Equation 1:
y + 2(1) - 1 = 0
y + 2 - 1 = 0
y + 1 = 0
y = -1
So the solution to the simultaneous equations is x = 1 and y = -1.