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Chapter 8: Problem 5

Solve each system of linear equations by substitution. $$\begin{aligned} &3 x+y=5\\\ &2 x-5 y=-8 \end{aligned}$$

Short Answer

Expert verified
The solution is \((x, y) = (1, 2)\).

Step by step solution

01

Solve for y in the first equation

The first equation is \( 3x + y = 5 \). To solve for \( y \), subtract \( 3x \) from both sides: \( y = 5 - 3x \). This expression for \( y \) will be substituted into the second equation.
02

Substitute the expression for y into the second equation

Take the expression for \( y \) from Step 1 and substitute it into the second equation, \( 2x - 5y = -8 \). Replace \( y \) with \( 5 - 3x \): \( 2x - 5(5 - 3x) = -8 \).
03

Simplify and solve for x

First, distribute \(-5\) into \( (5 - 3x) \), resulting in \(-25 + 15x\). The equation becomes \( 2x + 15x - 25 = -8 \). Combine like terms: \( 17x - 25 = -8 \). Add \( 25 \) to both sides to get \( 17x = 17 \). Divide both sides by \( 17 \) to find \( x = 1 \).
04

Substitute x back into the expression for y

With \( x = 1 \), substitute back into the expression for \( y \) found in Step 1: \( y = 5 - 3(1) \). Simplifying gives \( y = 5 - 3 = 2 \).
05

Verify the solution

Substitute \( x = 1 \) and \( y = 2 \) back into the original equations to verify the solution. First equation: \( 3(1) + 2 = 5 \), which is true. Second equation: \( 2(1) - 5(2) = -8 \), which is also true. Thus, the solution \( (x, y) = (1, 2) \) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful technique for solving systems of linear equations. It involves expressing one variable in terms of the other, then substituting that expression into the other equation. This simplifies the two-variable problem into a single equation with one unknown, making it easier to solve.To effectively use the substitution method:
  • Start by isolating one of the variables in one of the equations. Often, you'll want to choose the equation where isolating a variable seems straightforward. In our example, we isolated \( y \) in the first equation, resulting in \( y = 5 - 3x \).
  • Next, substitute this expression into the other equation. This replacement helps eliminate one variable, allowing you to focus on solving for the remaining one.
With practice, using substitution becomes a valuable tool to effortlessly solve linear equations, especially when combined with methods like elimination for more complicated systems.
Solving Equations
Solving equations is a fundamental part of algebra. It involves finding the value of variables that make the equation true. With linear equations, this process typically includes straightforward steps like isolating the variable, combining like terms, and using basic arithmetic operations.In the step-by-step solution, the equation \( 2x - 5(5 - 3x) = -8 \) was tackled by:
  • Distributing to eliminate parentheses: The distribution rule ensures that every term inside parentheses is multiplied by the factor outside, leading to \(-25 + 15x \).
  • Combining like terms: This consolidates similar expressions, simplifying to \( 17x - 25 = -8 \).
  • Undoing operations: Solving for \( x \) means reversing the steps used to form the equation. This led to isolating \( 17x \) and then dividing by 17 to find \( x = 1 \).
When you solve equations, remember to check your work by substituting the found values back into the original equations to verify accuracy.
Linear Algebra
Linear algebra encompasses the study of lines and planes, which are represented through systems of linear equations. It's a vital area of mathematics with applications in numerous fields like science, engineering, and economics. What makes linear algebra exciting is its capacity to describe multidimensional spaces using simple equations. Each linear equation represents a line in a plane, and systems of these equations can intersect at points, creating solutions:
  • A unique solution, which is the point of intersection of lines, like the point \( (x, y) = (1, 2) \) in our example.
  • No solution, when lines are parallel and never intersect.
  • Infinite solutions, when lines are coincident and overlap entirely.
By delving into linear algebra, you gain insight into the geometric implications of solving these systems, paving the way towards understanding more complex mathematical concepts.

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Most popular questions from this chapter

$$\begin{aligned} &\text { Verify that } A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{rr}d & -b \\\\-c & a \end{array}\right] \text { is the inverse of }\\\ &A=\left[\begin{array}{ll}a & b \\\c & d\end{array}\right], \text { provided } a d-b c \neq 0 \end{aligned}$$

You are asked to model a set of three points with a quadratic function \(y=a x^{2}+b x+c\) and determine the quadratic function. a. Set up a system of equations, use a graphing utility or graphing calculator to solve the system by entering the coefficients of the augmented matrix. b. Use the graphing calculator commands \([\text { STAT }]\) QuadReg to model the data using a quadratic function. Round your answers to two decimal places. $$(-9,20),(2,-18),(11,16)$$

In calculus, the first steps when solving the problem of finding the area enclosed by a set of curves are similar to those for finding the feasible region in a linear programming problem. In Exercises \(95-98,\) graph the system of inequalities and identify the vertices, that is, the points of intersection of the given curves. $$\begin{aligned} &\begin{array}{l} y \leq x^{2} \\ y \geq x^{3} \end{array}\\\ &x \geq 0 \end{aligned}$$

Maximize the objective function \(z=x+2 y\) subject to the conditions, where \(a > b > 0\). $$\begin{aligned}x+y & \geq a \\\\-x+y & \leq a \\\x+y & \leq a+b \\\\-x+y & \geq a-b\end{aligned}$$

In Exercises \(79-84,\) determine whether each statement is true or false. An objective function subject to constraints that correspond to a bounded region always has a maximum and a minimum.
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