Use the method of substitution to solve the system.
xy=1
4x-y+3=0
xy = 1,
4x - y + 3 = 0.
Solve 1st Eq for x:
X = 1/y,
Substitute 1/y for x in EQ2:
4/y - y + 3 = 0,
Multiply both sides by -y:
-4 + y^2 - 3y = 0,
y^2 - 3y - 4 = 0,
Factor the trinomial:
(y + 1) (y - 4) = 0,
y + 1 = 0,
y = -1.
y - 4 = 0,
y = 4.
Substitute -1 for y in Eq1:
x*-1 = 1,
x = -1.
Substitute 4 for y in Eq1:
x * 4 = 1,
x = 1/4.
Solution set: (-1 , -1) , (1/4 , 4).
To solve the system of equations using the method of substitution, we need to solve one equation for one variable and substitute that expression into the other equation.
Let's solve the first equation, xy=1, for y:
xy=1
Divide both sides by x:
y = 1/x
Now, we can substitute this expression for y in the second equation:
4x - y + 3 = 0
4x - (1/x) + 3 = 0
To simplify this equation, let's find a common denominator and combine like terms:
Multiply each term by x to get rid of the denominators:
4x^2 - 1 + 3x = 0
Combine like terms:
4x^2 + 3x - 1 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use factoring:
The equation can be factored as:
(2x - 1)(2x + 1) = 0
This gives us two possible solutions:
2x - 1 = 0 or 2x + 1 = 0
For the first equation:
2x - 1 = 0
Add 1 to both sides:
2x = 1
Divide both sides by 2:
x = 1/2
For the second equation:
2x + 1 = 0
Subtract 1 from both sides:
2x = -1
Divide both sides by 2:
x = -1/2
So, the solutions to the system of equations are x = 1/2 and x = -1/2.
Now, we can substitute these values back into the original equation to find the corresponding y-values:
For x = 1/2:
xy = 1
(1/2)y = 1
y = 2
For x = -1/2:
xy = 1
(-1/2)y = 1
y = -2
Therefore, the solutions to the system of equations are:
x = 1/2, y = 2
x = -1/2, y = -2
To solve the system using the method of substitution, we can use one equation to solve for one variable, and then substitute that expression into the other equation. Let's start with the first equation:
xy = 1
We can rearrange this equation to solve for one variable. Let's solve for y:
y = 1/x
Now, we can substitute this expression for y into the second equation:
4x - (1/x) + 3 = 0
To simplify this equation, we can multiply through by x to eliminate the fraction:
4x^2 - 1 + 3x = 0
Now, we have a quadratic equation. Rearranging the terms:
4x^2 + 3x - 1 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In our equation, a = 4, b = 3, and c = -1. Substituting these values into the quadratic formula:
x = (-(3) ± √((3)^2 - 4(4)(-1)))/(2(4))
Simplifying further:
x = (-3 ± √(9 + 16))/(8)
x = (-3 ± √25)/(8)
x = (-3 ± 5)/(8)
Now, we have two possible values for x:
1. x = (-3 + 5)/8 = 2/8 = 1/4
2. x = (-3 - 5)/8 = -8/8 = -1
Now, we substitute each value of x back into the equation y = 1/x to find the corresponding values of y:
For x = 1/4:
y = 1/(1/4) = 4/1 = 4
For x = -1:
y = 1/(-1) = -1
Therefore, the two solutions to the system of equations are (1/4, 4) and (-1, -1).