By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.
<h3>How to solve a system of equations</h3>
In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:
x - y = 5 (1)
x² · y = 5 · x + 6 (2)
By (1):
y = x + 5
By substituting on (2):
x² · (x + 5) = 5 · x + 6
x³ + 5 · x² - 5 · x - 6 = 0
(x + 5.693) · (x - 1.430) · (x + 0.737) = 0
There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737
And the y-values are found by evaluating on (1):
y = x + 5
x₁ ≈ 5.693
y₁ ≈ 10.693
x₂ ≈ 1.430
y₂ ≈ 6.430
x₃ ≈ - 0.737
y₃ ≈ 4.263
By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.
To learn more on nonlinear equations: brainly.com/question/20242917
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Answer:
255b
Step-by-step explanation:
subtract 256b by b
Answer: The Last One
Step-by-step explanation: The Last One Would Be A Product Of Negative
Answer:
y=3/5*x, A(0,0), B(1,3/5), C(2,6/5)
Step-by-step explanation:
A(5,3)=(x1,y1)....x1=5, y1=3
m=3/5
y-y1=m(x-x1)
y-3=3/5(x-5)
y-3=3/5*x-3/5*5
y-3=3/5*x - 3
y=3/5*x-3+3
y=3/5*x
x=0, y=3/5*x, y=3/5*0=0 A(0,0)
x=1, y=3/5*1=3/5 B(1,3/5)
x=2, y=3/5*2=6/5 C(2,6/5)
If square roots includes exponents then that one is first. but if not then it is division.
