/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Solve the system using any metho... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the system using any method. $$ \begin{array}{l} 2 x-7 y=2400 \\ -4 x+1800=y \end{array} $$

Short Answer

Expert verified
The solution is \(x = 500\) and \(y = -200\).

Step by step solution

01

- Write down the equations

The system of equations is: 1. \(2x - 7y = 2400\) 2. \(-4x + 1800 = y\)
02

- Substitute Equation 2 into Equation 1

Use the second equation \(y = -4x + 1800\) and substitute this expression for y in the first equation: \[2x - 7(-4x + 1800) = 2400\]
03

- Simplify the equation

Distribute and combine like terms: \[2x + 28x - 12600 = 2400\] \[30x - 12600 = 2400\]
04

- Solve for x

Isolate \(x\) by adding 12600 to both sides and then dividing by 30: \[30x = 15000\] \[x = 500\]
05

- Solve for y

Substitute \(x = 500\) back into the second equation to solve for y: \[y = -4(500) + 1800\] \[y = -2000 + 1800\] \[y = -200\]
06

- Solution verification

Verify the solution by substituting \(x = 500\) and \(y = -200\) into the original equations to ensure both are satisfied.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

substitution method
The substitution method is a key technique for solving systems of linear equations. In this method, one of the equations is solved for one variable in terms of the other variable. This expression is then substituted into the other equation. This helps in reducing the system to a single equation with one variable.
In our example, we have two equations:
1. \(2x - 7y = 2400\)
2. \(-4x + 1800 = y\)
We solve the second equation for y: \(y = -4x + 1800\).
Next, we substitute this expression into the first equation to solve for x.
Benefits of the substitution method:
  • It simplifies the process by reducing the number of variables.
  • It is straightforward and systematic.
  • It is helpful when one equation is already solved for a variable.
linear equations
Linear equations form the backbone of algebra. They are equations that make a straight line when graphed. Each term in a linear equation is either a constant or the product of a constant and a single variable.
In our problem, we have two linear equations:
\(2x - 7y = 2400\)
\(-4x + 1800 = y\)
These equations describe relationships between the variables x and y. They are called 'linear' because their graph will be a straight line.
Understanding the structure of linear equations:
  • They typically have constants and variables.
  • Each term is of the first degree (no exponents other than 1).
  • They can have multiple variables but with no terms that are products of variables.
algebraic manipulation
Algebraic manipulation involves rearranging equations and expressions to isolate a variable or simplify the equation. This is frequently used in solving systems of equations.
In our step-by-step solution, we performed various algebraic manipulation techniques:
Step 3: Simplified the substituted equation \(2x - 7(-4x + 1800) = 2400\) to \(2x + 28x - 12600 = 2400\).
Step 4: Solved for x:
\(30x - 12600 = 2400\)
Added 12600 to both sides:
\(30x = 15000\)
Finally, divided by 30 to find:
\(x = 500\)
The essence of algebraic manipulation is dealing with equations through:
  • Combining like terms.
  • Distributing to remove parentheses.
  • Addition or subtraction of the same quantity to both sides.
  • Division or multiplication by a nonzero number to isolate the variable.
solution verification
Verification confirms that our solution satisfies the original equations. This step ensures the accuracy of our solution.
In our example, after solving for \(x = 500\) and \(y = -200\), we plug these values back into the original equations:
First equation: \(2(500) - 7(-200) = 2400\)
Simplifies to:
\(1000 + 1400 = 2400\)
which is true.
Second equation: \(-4(500) + 1800 = -200\)
Simplifies to:
\(-2000 + 1800 = -200\)
which is true.
Verification steps:
  • Substitute the found values into each original equation.
  • Confirm the left-hand side equals the right-hand side for both equations.
  • If both equations hold true, the solution is correct.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A patient undergoing a heart scan is given a sample of fluorine- \(18\left({ }^{18} \mathrm{~F}\right)\). After \(4 \mathrm{hr}\), the radioactivity level in the patient is \(44.1 \mathrm{MBq}\) (megabecquerel). After \(5 \mathrm{hr}\), the radioactivity level drops to \(30.2 \mathrm{MBq}\). The radioactivity level \(Q(t)\) can be approximated by \(Q(t)=Q_{0} e^{-k t},\) where \(t\) is the time in hours after the initial dose \(Q_{0}\) is administered. a. Determine the value of \(k\). Round to 4 decimal places. b. Determine the initial dose, \(Q_{0}\). Round to the nearest whole unit. c. Determine the radioactivity level after \(12 \mathrm{hr}\). Round to 1 decimal place.

Explain how test points are used to determine the region of the plane that represents the solution to an inequality in two variables.

The attending physician in an emergency room treats an unconscious patient suspected of a drug overdose. The physician does not know the initial concentration \(A_{0}\) of the drug in the bloodstream at the time of injection. However, the physician knows that after \(3 \mathrm{hr}\), the drug concentration in the blood is \(0.69 \mu \mathrm{g} / \mathrm{dL}\) and after \(4 \mathrm{hr}\), the concentration is \(0.655 \mu \mathrm{g} / \mathrm{dL}\). The model \(A(t)=A_{0} e^{-k t}\) represents the drug concentration \(A(t)\) (in \(\mu \mathrm{g} / \mathrm{dL}\) ) in the bloodstream \(t\) hours after injection. The value of \(k\) is a constant related to the rate at which the drug is removed by the body. a. Substitute 0.69 for \(A(t)\) and 3 for \(t\) in the model and write the resulting equation. b. Substitute 0.655 for \(A(t)\) and 4 for \(t\) in the model and write the resulting equation. c. Use the system of equations from parts (a) and (b) to solve for \(k .\) Round to 3 decimal places. d. Use the system of equations from parts (a) and (b) to approximate the initial concentration \(A_{0}\) (in \(\mu \mathrm{g} / \mathrm{dL}\) ) at the time of injection. Round to 2 decimal places. e. Determine the concentration of the drug after \(12 \mathrm{hr}\). Round to 2 decimal places.

Use a graphing utility to approximate the solution(s) to the system of equations. Round the coordinates to 3 decimal places. $$ \begin{array}{l} y=x^{2}-8 x+20 \\ y=4 \log x \end{array} $$

Juan borrows \(\$ 100,000\) to pay for medical school. He borrows part of the money from the school whereby he will pay \(4.5 \%\) simple interest. He borrows the rest of the money through a government loan that will charge him \(6 \%\) interest. In both cases, he is not required to pay off the principal or interest during his 4 yr of medical school. However, at the end of \(4 \mathrm{yr}\), he will owe a total of \(\$ 19,200\) for the interest from both loans. How much did he borrow from each source?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.