Solve the System of Equations x+y=-13 and 4x+6y=-72
The problem you've been given is a request to find the values of the variables x and y that satisfy both of the two provided linear equations simultaneously. These equations form a system that is typically solved using methods such as substitution, elimination, or matrix operations (if dealing with more complex systems). The first equation is a simple linear equation with two variables, x and y. The second equation is another linear equation that, together with the first one, must be true for the same values of x and y. The goal is to determine the exact numerical values for x and y that make both of these statements true at the same time.
$x + y = - 13$and$4 x + 6 y = - 72$
Answer
Solution:
Step:1
Isolate $x$ in the first equation by moving $y$ to the other side: $x = -13 - y$. Keep the second equation as is: $4x + 6y = -72$.
Step:2
Substitute $-13 - y$ for $x$ in the second equation.
Step:2.1
Substitute $x$ in $4x + 6y = -72$ with $-13 - y$: $4(-13 - y) + 6y = -72$. Keep the first equation unchanged: $x = -13 - y$.
Step:2.2
Begin simplifying the equation.
Step:2.2.1
Start by expanding $4(-13 - y) + 6y$.
Step:2.2.1.1
Distribute $4$ across $-13$ and $-y$.
Step:2.2.1.1.1
Use the distributive property: $4 \cdot -13 + 4 \cdot (-y) + 6y = -72$. The first equation remains: $x = -13 - y$.
Step:2.2.1.1.2
Calculate $4 \cdot -13$: $-52 + 4 \cdot (-y) + 6y = -72$. The first equation remains: $x = -13 - y$.
Step:2.2.1.1.3
Multiply $-1$ by $4$ to get $-4y$: $-52 - 4y + 6y = -72$. The first equation remains: $x = -13 - y$.
Step:2.2.1.2
Combine like terms $-4y$ and $6y$: $-52 + 2y = -72$. The first equation remains: $x = -13 - y$.
Step:3
Solve for $y$ in the equation $-52 + 2y = -72$.
Step:3.1
Move the constant term to the opposite side of the equation.
Step:3.1.1
Add $52$ to both sides: $2y = -72 + 52$. The first equation remains: $x = -13 - y$.
Step:3.1.2
Combine $-72$ and $52$: $2y = -20$. The first equation remains: $x = -13 - y$.
Step:3.2
Divide the equation by $2$ to solve for $y$.
Step:3.2.1
Divide both sides by $2$: $\frac{2y}{2} = \frac{-20}{2}$. The first equation remains: $x = -13 - y$.
Step:3.2.2
Simplify both sides.
Step:3.2.2.1
Cancel the common factor of $2$: $y = \frac{-20}{2}$. The first equation remains: $x = -13 - y$.
Step:3.2.3
Divide $-20$ by $2$: $y = -10$. The first equation remains: $x = -13 - y$.
Step:4
Substitute $-10$ for $y$ in the first equation to find $x$.
Step:4.1
Substitute $y$ in $x = -13 - y$ with $-10$: $x = -13 - (-10)$. Keep the value of $y$: $y = -10$.
Step:4.2
Simplify the equation.
Step:4.2.1
Simplify $-13 - (-10)$.
Step:4.2.1.1
Multiply $-1$ by $-10$: $x = -13 + 10$. Keep the value of $y$: $y = -10$.
Step:4.2.1.2
Add $-13$ and $10$: $x = -3$. Keep the value of $y$: $y = -10$.
Step:5
The solution to the system is the set of values for $x$ and $y$ that satisfy both equations: $(-3, -10)$.
Step:6
The solution can be expressed in different formats.
Point Form: $(-3, -10)$
Equation Form: $x = -3, y = -10$
Step:7
End of the solution process.
Knowledge Notes:
To solve a system of linear equations, one can use substitution or elimination methods. The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation. This process reduces the system to a single equation with one variable, which can be solved using algebraic techniques.
The steps involved in the substitution method include:
Isolate one variable in one of the equations.
Substitute the expression for the isolated variable into the other equation.
Simplify the resulting equation and solve for the remaining variable.
Substitute the found value back into one of the original equations to find the value of the other variable.
Check the solution by plugging the values into both original equations to ensure they are satisfied.
In this problem, the distributive property is used to expand expressions, and like terms are combined to simplify the equations. The distributive property states that $a(b + c) = ab + ac$. When solving for a variable, it is important to perform the same operation on both sides of the equation to maintain equality. The solution to the system is an ordered pair that represents the point of intersection of the two lines represented by the equations in a Cartesian plane.
