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Chapter 6: Problem 19

Solve each system by substitution. $$\begin{aligned}&4 x-5 y=-11\\\&2 x+y=5\end{aligned}$$

Short Answer

Expert verified
The solution is \((x, y) = (1, 3)\).

Step by step solution

01

Solve One Equation for One Variable

Let's solve the second equation for \( y \). We have:\[2x + y = 5\]Rearranging for \( y \) gives:\[y = 5 - 2x\]
02

Substitute and Solve

Substitute \( y = 5 - 2x \) into the first equation. The first equation is:\[4x - 5y = -11\]Replacing \( y \) with \( 5 - 2x \) yields:\[4x - 5(5 - 2x) = -11\]Simplify and solve for \( x \):\[4x - 25 + 10x = -11\]\[14x - 25 = -11\]Add 25 to both sides:\[14x = 14\]Divide by 14:\[x = 1\]
03

Substitute Back and Solve for Second Variable

Now that we know \( x = 1 \), substitute back into the equation for \( y \):\[y = 5 - 2(1)\]Calculate \( y \):\[y = 5 - 2 = 3\]
04

Check the Solution in Both Equations

Verify the solution \((1, 3)\) satisfies the original equations.For the first equation:\[4(1) - 5(3) = -11 \]\[4 - 15 = -11 \]This is true.For the second equation:\[2(1) + 3 = 5 \]\[2 + 3 = 5 \]This is also true.

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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 popular technique for solving systems of linear equations. The idea is to solve one equation for one of the variables, and then substitute the expression from this equation into the other equation. This way, you reduce the system to a single equation with one variable, which can be easily solved.

Here's a step-by-step idea of how the method works:
  • Solve one equation for one variable: Choose one of the equations and solve for either variable. This usually involves addition, subtraction, multiplication, or division to isolate the variable on one side of the equation.
  • Substitute the expression into the other equation: Replace the chosen variable in the other equation with the expression you found in the first step. This gives you a single equation with one variable.
  • Solve for the second variable: Once you substitute, you'll have an equation that can be solved like a regular algebraic equation to find the value of the second variable.
  • Substitute back to find the first variable: Use the value of the second variable and substitute it back into any equation (usually the one you rearranged) to find the first variable.
  • Check your solutions: Always plug the variables back into the original equations to ensure they satisfy both equations.
This method is especially useful when one of the equations can be easily manipulated to express one variable in terms of the other.
Linear Equations
Linear equations are mathematical equations that represent straight lines when plotted on a graph. They have many practical applications, especially in fields like economics, physics, and engineering. A typical linear equation in two variables is written as: \[ ax + by = c \] where \( a \), \( b \), and \( c \) are constants.

Key aspects of linear equations include:
  • Variables: Usually, linear equations involve two variables, commonly represented as \( x \) and \( y \). The solution to a system of linear equations is a pair \((x, y)\) that satisfies all the equations in the system.
  • Coefficients and Constant Terms: Coefficients are the numerical factors of the variables. The constant term is the number that doesn鈥檛 change and stands alone.
  • Graphing: When graphed, linear equations form a straight line. The slope of the line represents the ratio of change between the two variables, and the y-intercept is where the line crosses the y-axis.
  • Systems of Linear Equations: Systems involve multiple linear equations. The solution is the point where the lines intersect, which corresponds to the values for the variables that satisfy all the equations simultaneously.
Understanding linear equations is essential as they form the foundation for more advanced topics in algebra and calculus.
Algebraic Solutions
Algebraic solutions involve solving equations using algebraic manipulations. This includes techniques like substitution, elimination, and rearrangement to isolate variables and solve for their values.

In the context of the problem:
  • Substituting expressions: Using substitution allows you to replace a variable with its equivalent expression obtained from another equation, simplifying the problem.
  • Simplifying expressions: By systematically simplifying each step, such as combining like terms and isolating variables, one can eventually solve for the unknowns.
  • Verification: After obtaining a solution, it's crucial to verify that the values satisfy the original equations. This can involve simply plugging the numbers back into the equations to see if true statements result.
  • Systematic Approach: Algebra provides a structured method for solving complex problems by breaking them down into more manageable parts and solving each part at a time.
Algebraic solutions require careful handling of operations to ensure accuracy and solve systems effectively.

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Most popular questions from this chapter

Explain how one can determine whether a system is inconsistent or has dependent equations when using the substitution or elimination method.

Draw a sketch of the two graphs described with the indicated number of points of intersection. (There may be more than one way to do this.) A line and a circle; two points.

The relationship between a professional basketball player's height \(h\) in inches and weight \(w\) in pounds was modeled by using two samples of players. The resulting equations were $$\begin{aligned}&w=7.46 h-374\\\&w=7.93 h-405\end{aligned}$$ and Assume that \(65 \leq h \leq 85\) (a) Use each equation to predict the weight to the nearest pound of a professional basketball player who is 6 feet 11 inches. (b) Determine graphically the height at which the two models give the same weight. (c) For each model, what change in weight is associated with a 1 -inch increase in height?

Draw a sketch of the two graphs described with the indicated number of points of intersection. (There may be more than one way to do this.) A circle and a parabola; four points.

In certain parts of the Rocky Mountains, deer are the main food source for mountain lions. When the deer population \(d\) is large, the mountain lions ( \(m\) ) thrive. However, a large mountain lion population drives down the size of the deer population. Suppose the fluctuations of the two populations from year to year can be modeled with the matrix equation $$\left[\begin{array}{c} m_{n+1} \\ d_{n+1} \end{array}\right]=\left[\begin{array}{cc} 0.51 & 0.4 \\ -0.05 & 1.05 \end{array}\right]\left[\begin{array}{l} m_{n} \\ d_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of animals in the two populations after \(n\) years and \(n+1\) years, where the number of deer is measured in hundreds. (a) Give the equation for \(d_{n+1}\) obtained from the second row of the square matrix. Use this equation to determine the rate the deer population will grow from year to year if there are no mountain lions. (b) Suppose we start with a mountain lion population of 2000 and a deer population of \(500,000\) (that is, 5000 hundred deer). How large would each population be after 1 year? 2 years? (c) Consider part (b), but change the initial mountain lion population to \(4000 .\) Show that the populations would both grow at a steady annual rate of \(1 \$ 6\).
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