/*! 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 25 Solve by substitution. \(4 x+3... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve by substitution. \(4 x+3 y=0\) \(2 x-y=0\)

Short Answer

Expert verified
The solution to the system of equations is \(x = 0\) and \(y = 0\).

Step by step solution

01

Solve one of the equations for one variable

The second equation is simpler, so let's solve it for \(y\). From the equation \(2x - y = 0\), we rearrange to find \(y = 2x\)
02

Substitue this solution into the another equation

Substitute \(y = 2x\) into the first equation \(4x + 3y = 0\). We get \(4x + 3(2x) = 0\), which simplifies to \(4x + 6x = 0\), and then to \(10x = 0\)
03

Solve for the unknown variable

Solve for \(x\) by dividing both sides by 10, we get \(x = 0\)
04

Determine the other variable

Substitute \(x = 0\) into one of the original equations to solve for \(y\). Choosing the equation \(2x - y = 0\), and substituting, we get \(0 - y = 0\), which simplifies to \(y = 0\)

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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 one of the strategies used to solve systems of algebraic equations. This technique involves expressing one variable in terms of another from one equation and substituting this expression into another equation. This helps to reduce a system of equations to a single equation in one variable, which can then be solved using basic algebraic rules.

Here's how the substitution method works in practice:
  • Solve one of the equations for one of its variables.
  • Substitute the expression obtained in step 1 into the other equation.
  • Solve the resulting single-variable equation.
  • Back-substitute the found value into one of the original equations to determine the value of the other variable.
In our example, we chose the second equation to solve for y, because it was simpler and led to a straightforward expression of y in terms of x. Substituting this expression into the first equation simplified the system to one equation with one unknown, which was easy to solve.
Algebraic Equations
Algebraic equations are mathematical statements that show the equality of two expressions with one or more unknowns (variables). They are foundational in algebra and form the basis for solving various problems. Equations come in different forms - linear, quadratic, polynomial, and others - each with unique properties and methods of solution.

In the context of our exercise, we dealt with a system of linear algebraic equations. These equations are recognizable by the variables appearing to the first power only and being arranged in a straight line when graphed. By solving algebraic equations, we can find the values of the unknown variables that satisfy all equations simultaneously.
Solving Linear Equations
Solving linear equations involves finding the value of the variable that makes the equation true. Linear equations are simple to solve and usually require a few basic steps:
  • Isolate the variable on one side of the equation.
  • Simplify the equation by combining like terms and reducing fractions if necessary.
  • Check the solution by substituting it back into the original equation to ensure it satisfies the equation.
When solving systems of linear equations, multiple methods, including substitution, elimination, and graphing, can be used. In this exercise, we used the substitution method, which reduced our system to a single linear equation that was straightforward to solve. After finding the value of x, we back-substituted into the original equations to find the value of y, hence solving the entire system.

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Most popular questions from this chapter

Ten coworkers formed an investment club, and each member deposited \(\$ 2000\) in the club's account. The club decided to take the total amount and invest some of it in preferred stock that pays \(8 \%\) annual simple interest and the remainder in a municipal bond that pays \(7 \%\) annual simple interest. The amount of interest earned each year from the investments was \(\$ 1520 .\) How much was invested in each?

\(\quad\) A baker purchased 12 Ib of wheat flour and 15 lb of rye flour for a total cost of \(\$ 18.30 .\) A second purchase, at the same prices, included 15 lb of wheat flour and 10 Ib of rye flour. The cost of the second purchase was \(\$ 16.75 .\) Find the cost per pound of the wheat flour and of the rye flour.

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A mortgage broker purchased two trust deeds for a total of \(\$ 250,000 .\) One trust deed earns \(7 \%\) simple annual interest, and the second earns \(8 \%\) simple annual interest. If the total annual interest earned from the two trust deeds is \(\$ 18,500,\) what was the purchase price of each trust deed?

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