Solve the simultaneous equations giving your answers to 3 s.f if necessary
In(x-y) - 0
2 In x = ln (y+4)
We have the following simultaneous equations:
1) In(x-y) = 0
2) 2 In x = ln (y+4)
To solve this system of equations, we'll start with equation 1 and solve for x-y.
1) In(x-y) = 0
We know that ln(1) = 0, so x-y = 1.
Now, let's substitute x-y = 1 into equation 2:
2 In x = ln (y+4)
2 ln x = ln (y+4)
ln x^2 = ln (y+4)
Since the natural logarithm is a one-to-one function, we can equate the arguments:
x^2 = y+4
Now, we have a system of equations:
x - y = 1
x^2 - y = 4
We can solve this system by substitution or elimination method.
Using substitution, we can solve equation 1 for x and substitute into equation 2:
x = y + 1
Substituting into equation 2:
(y + 1)^2 - y = 4
y^2 + 2y + 1 - y = 4
y^2 + y - 3 = 0
We can solve this quadratic equation for y by factoring or using the quadratic formula.
(y - 1)(y + 3) = 0
This gives us two possible values for y: y = 1 or y = -3.
Case 1: y = 1
Substituting y = 1 into x - y = 1:
x - 1 = 1
x = 2
Case 2: y = -3
Substituting y = -3 into x - y = 1:
x - (-3) = 1
x + 3 = 1
x = -2
The solutions to the system of equations are (x, y) = (2, 1) and (-2, -3) rounded to 3 significant figures if necessary.
To solve the simultaneous equations, we will first rewrite the equations using the natural logarithm rules:
Equation 1: ln(x - y) = 0
Equation 2: 2ln(x) = ln(y + 4)
Now, we can solve them step-by-step.
Step 1: Solve Equation 1.
ln(x - y) = 0
To remove the natural logarithm, we can exponentiate both sides using e as the base (e is Euler's number, approximately 2.71828).
e^(ln(x - y)) = e^0
Simplifying,
x - y = 1
Step 2: Solve Equation 2.
2ln(x) = ln(y + 4)
Using the property of logarithms, we can rewrite the equation as:
ln(x^2) = ln(y + 4)
Expanding the logarithm,
x^2 = y + 4
Step 3: Use the equation x - y = 1 and x^2 = y + 4 to solve for x and y.
Substitute x - y = 1 into x^2 = y + 4.
(x - y)^2 = y + 4
Expanding,
x^2 - 2xy + y^2 = y + 4
Since we have another equation x^2 = y + 4, we can substitute this into the equation above.
(y + 4) - 2xy + y^2 = y + 4
Simplifying further,
y^2 - 2xy - y = 0
Factoring out y,
y(y - 1) - 2x(y - 1) = 0
(y - 1)(y - 2x) = 0
From here, we have two cases to consider:
Case 1: y - 1 = 0
y = 1
Substitute y = 1 back into the equation x^2 = y + 4.
x^2 = 1 + 4
x^2 = 5
x ≈ 2.236 (rounded to 3 significant figures)
Case 2: y - 2x = 0
y = 2x
Substitute y = 2x into the equation x^2 = y + 4.
x^2 = 2x + 4
Rearranging,
x^2 - 2x - 4 = 0
Using the quadratic formula,
x = (-(-2) ± √((-2)^2 - 4(1)(-4))) / (2(1))
x = (2 ± √(4 + 16)) / 2
x = (2 ± √20) / 2
x = (2 ± 2√5) / 2
x = 1 ± √5
So, the solutions to the simultaneous equations, rounded to 3 significant figures, are:
x ≈ 2.236 and y ≈ 1
x ≈ -0.236 and y ≈ -0.472