Chapter 4: Q 63. (page 387)

In the following exercise, solve the systems of equations by elimination.
$\{\begin{cases} 4 x + 3 y = 2 \\ 20 x + 15 y = 10 \end{cases}$

Short Answer

Expert verified

The given system has infinitely many solutions.

Step by step solution

Step 1. Given Information

The given linear equations are :

$\{\begin{cases} 4 x + 3 y = 2 \\ 20 x + 15 y = 10 \end{cases}$

Step 2. Calculating the value of y

To calculate the value of $y$ , we have to eliminate the $x$ term. For which we will do the following changes in the equation, so that $x$ term can be easily eliminated.

Multiply the first equation by $- 5$ ,

localid="1644751669758" $- 5 (4 x + 3 y = 2) - 20 x - 15 y = - 10$

now, the equations are,

$- 20 x - 15 y = - 10 20 x + 15 y = 10$

now simply add the equations

$- 20 x - 15 y = - 10 20 x + 15 y = 10 0 = 0$

this means that the above equation is satisfied for all the values of $y$ and $x$ .

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In the following exercise, solve the systems of equations by elimination.
$\{\begin{cases} 4 x + 3 y = 2 \\ 20 x + 15 y = 10 \end{cases}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculating the value of y

One App. One Place for Learning.

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In the following exercise, solve the systems of equations by elimination.4x+3y=220x+15y=10

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculating the value of y

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Calculus

Applied Mathematics

Statistics

Geometry

Probability and Statistics

Study anywhere. Anytime. Across all devices.

In the following exercise, solve the systems of equations by elimination.
$\{\begin{cases} 4 x + 3 y = 2 \\ 20 x + 15 y = 10 \end{cases}$