Solve the System of Equations x^2+y^2=10 y=2x-5
The problem presents a system of equations that consists of two equations with two variables, x and y. The first equation, x^2 + y^2 = 10, represents a circle with a certain radius centered at the origin on a coordinate plane. The second equation, y = 2x - 5, represents a straight line. The question asks to find the point(s) where the circle and the line intersect by solving this system simultaneously.
$x^{2} + y^{2} = 10$$y = 2 x - 5$
Answer
Solution:
Step:1
Substitute $2x - 5$ for $y$ in the equation $x^2 + y^2 = 10$.
Step:1.1
Substitute $y$ in $x^2 + y^2 = 10$ with $2x - 5$.
$x^2 + (2x - 5)^2 = 10$ $y = 2x - 5$
Step:1.2
Expand and simplify the equation.
Step:1.2.1
Expand $x^2 + (2x - 5)^2$.
Step:1.2.1.1
Expand each term individually.
Step:1.2.1.1.1
Rewrite $(2x - 5)^2$ as $(2x - 5)(2x - 5)$.
$x^2 + (2x - 5)(2x - 5) = 10$ $y = 2x - 5$
Step:1.2.1.1.2
Use the FOIL Method to expand $(2x - 5)(2x - 5)$.
Step:1.2.1.1.2.1
Distribute $2x$ across $(2x - 5)$ and $-5$ across $(2x - 5)$.
$x^2 + 2x(2x - 5) - 5(2x - 5) = 10$ $y = 2x - 5$
Step:1.2.1.1.3
Combine like terms.
Step:1.2.1.1.3.1
Simplify and combine like terms.
$x^2 + 4x^2 - 10x - 10x + 25 = 10$ $y = 2x - 5$
Step:1.2.1.2
Combine $x^2$ and $4x^2$.
$5x^2 - 20x + 25 = 10$ $y = 2x - 5$
Step:2
Solve for $x$ in $5x^2 - 20x + 25 = 10$.
Step:2.1
Subtract $10$ from both sides.
$5x^2 - 20x + 15 = 0$ $y = 2x - 5$
Step:2.2
Factor the quadratic equation.
Step:2.3
Factor out the greatest common factor, which is $5$.
Step:2.3.1
Factor $5$ from each term.
$5(x^2 - 4x + 3) = 0$ $y = 2x - 5$
Step:2.3.2
Factor the trinomial $x^2 - 4x + 3$.
Step:2.3.2.1
Find two numbers that multiply to $3$ and add to $-4$.
$(x - 3)(x - 1) = 0$ $y = 2x - 5$
Step:2.4
Set each factor equal to zero and solve for $x$.
$x - 3 = 0$ or $x - 1 = 0$ $y = 2x - 5$
Step:2.5
Solve $x - 3 = 0$.
$x = 3$ $y = 2x - 5$
Step:2.6
Solve $x - 1 = 0$.
$x = 1$ $y = 2x - 5$
Step:2.7
The solutions for $x$ are $3$ and $1$.
$x = 3, 1$ $y = 2x - 5$
Step:3
Substitute $x = 3$ into $y = 2x - 5$ to find $y$.
$y = 2(3) - 5$
Step:3.1
Simplify to find $y$.
$y = 6 - 5$ $y = 1$ $x = 3$
Step:4
Substitute $x = 1$ into $y = 2x - 5$ to find $y$.
$y = 2(1) - 5$
Step:4.1
Simplify to find $y$.
$y = 2 - 5$ $y = -3$ $x = 1$
Step:5
The solution set is the pair of $x$ and $y$ values that satisfy both equations.
$(3, 1)$ and $(1, -3)$
Step:6
The solution can be presented in different formats.
Point Form: $(3, 1), (1, -3)$ Equation Form: $x = 3, y = 1$ and $x = 1, y = -3$
Knowledge Notes:
The problem involves solving a system of equations, one of which is a circle equation $x^2 + y^2 = 10$ and the other is a linear equation $y = 2x - 5$. The process includes substitution, simplification, expansion, factoring, and solving quadratic equations.
Substitution: Replace $y$ with $2x - 5$ in the circle equation to eliminate $y$ and obtain a single equation in terms of $x$.
Simplification: Simplify the equation by expanding and combining like terms.
Factoring: Factor the resulting quadratic equation to find the values of $x$ that satisfy the equation.
Solving Quadratic Equations: Set each factor equal to zero and solve for $x$.
Back Substitution: Substitute the found values of $x$ into the original linear equation to find the corresponding values of $y$.
Solution Set: The solution to the system of equations is the set of ordered pairs $(x, y)$ that satisfy both original equations.
The FOIL Method (First, Outer, Inner, Last) is used to expand binomials, and factoring involves finding two numbers that multiply to the constant term and add to the coefficient of the middle term in a quadratic equation. The distributive property is applied when expanding expressions, and the zero-product property is used to find the solutions of a factored equation.
