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Sagot :
To determine the nature of the system of linear equations [tex]\( 4x + 3y = 12 \)[/tex] and [tex]\( 2x + 9y = 15 \)[/tex], we need to analyze their solutions and the relationships between their coefficients.
### Step 1: Understanding the equations
We are given the following system of equations:
1. [tex]\( 4x + 3y = 12 \)[/tex]
2. [tex]\( 2x + 9y = 15 \)[/tex]
### Step 2: Solving the system using substitution or elimination
#### Using Substitution
Let's take the first equation and solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ y = \frac{12 - 4x}{3} \][/tex]
Now, substitute [tex]\( y \)[/tex] in the second equation:
[tex]\[ 2x + 9\left(\frac{12 - 4x}{3}\right) = 15 \][/tex]
Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x + 3(12 - 4x) = 15 \][/tex]
[tex]\[ 2x + 36 - 12x = 15 \][/tex]
[tex]\[ -10x + 36 = 15 \][/tex]
[tex]\[ -10x = 15 - 36 \][/tex]
[tex]\[ -10x = -21 \][/tex]
[tex]\[ x = \frac{21}{10} \][/tex]
Now substitute [tex]\( x = \frac{21}{10} \)[/tex] back into the expression for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{12 - 4\left(\frac{21}{10}\right)}{3} \][/tex]
[tex]\[ y = \frac{12 - \frac{84}{10}}{3} \][/tex]
[tex]\[ y = \frac{12 - 8.4}{3} \][/tex]
[tex]\[ y = \frac{3.6}{3} \][/tex]
[tex]\[ y = \frac{6}{5} \][/tex]
So the solution to the system of equations is [tex]\( x = \frac{21}{10} \)[/tex] and [tex]\( y = \frac{6}{5} \)[/tex].
### Step 3: Determining the nature of the lines
To determine if the lines are intersecting, parallel, or coincident, we need to look at the coefficients:
First equation: [tex]\( 4x + 3y = 12 \)[/tex]
- Coefficients: [tex]\( a_1 = 4 \)[/tex], [tex]\( b_1 = 3 \)[/tex], [tex]\( c_1 = -12 \)[/tex]
Second equation: [tex]\( 2x + 9y = 15 \)[/tex]
- Coefficients: [tex]\( a_2 = 2 \)[/tex], [tex]\( b_2 = 9 \)[/tex], [tex]\( c_2 = -15 \)[/tex]
### Step 4: Compare the coefficients
The criterion to determine the relationship:
- If [tex]\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)[/tex], the lines are intersecting.
- If [tex]\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \)[/tex] but [tex]\( \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \)[/tex], the lines are parallel.
- If [tex]\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)[/tex], the lines are coincident.
Evaluating the ratios:
[tex]\[ \frac{a_1}{a_2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ \frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3} \][/tex]
[tex]\[ \frac{c_1}{c_2} = \frac{-12}{-15} = \frac{4}{5} \][/tex]
Since [tex]\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)[/tex], the lines are intersecting.
### Conclusion
The system of equations [tex]\( 4x + 3y = 12 \)[/tex] and [tex]\( 2x + 9y = 15 \)[/tex] represents intersecting lines with the solution [tex]\( x = \frac{21}{10} \)[/tex] and [tex]\( y = \frac{6}{5} \)[/tex]. Thus, the lines are intersecting.
### Step 1: Understanding the equations
We are given the following system of equations:
1. [tex]\( 4x + 3y = 12 \)[/tex]
2. [tex]\( 2x + 9y = 15 \)[/tex]
### Step 2: Solving the system using substitution or elimination
#### Using Substitution
Let's take the first equation and solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ y = \frac{12 - 4x}{3} \][/tex]
Now, substitute [tex]\( y \)[/tex] in the second equation:
[tex]\[ 2x + 9\left(\frac{12 - 4x}{3}\right) = 15 \][/tex]
Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 2x + 3(12 - 4x) = 15 \][/tex]
[tex]\[ 2x + 36 - 12x = 15 \][/tex]
[tex]\[ -10x + 36 = 15 \][/tex]
[tex]\[ -10x = 15 - 36 \][/tex]
[tex]\[ -10x = -21 \][/tex]
[tex]\[ x = \frac{21}{10} \][/tex]
Now substitute [tex]\( x = \frac{21}{10} \)[/tex] back into the expression for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{12 - 4\left(\frac{21}{10}\right)}{3} \][/tex]
[tex]\[ y = \frac{12 - \frac{84}{10}}{3} \][/tex]
[tex]\[ y = \frac{12 - 8.4}{3} \][/tex]
[tex]\[ y = \frac{3.6}{3} \][/tex]
[tex]\[ y = \frac{6}{5} \][/tex]
So the solution to the system of equations is [tex]\( x = \frac{21}{10} \)[/tex] and [tex]\( y = \frac{6}{5} \)[/tex].
### Step 3: Determining the nature of the lines
To determine if the lines are intersecting, parallel, or coincident, we need to look at the coefficients:
First equation: [tex]\( 4x + 3y = 12 \)[/tex]
- Coefficients: [tex]\( a_1 = 4 \)[/tex], [tex]\( b_1 = 3 \)[/tex], [tex]\( c_1 = -12 \)[/tex]
Second equation: [tex]\( 2x + 9y = 15 \)[/tex]
- Coefficients: [tex]\( a_2 = 2 \)[/tex], [tex]\( b_2 = 9 \)[/tex], [tex]\( c_2 = -15 \)[/tex]
### Step 4: Compare the coefficients
The criterion to determine the relationship:
- If [tex]\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)[/tex], the lines are intersecting.
- If [tex]\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \)[/tex] but [tex]\( \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \)[/tex], the lines are parallel.
- If [tex]\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)[/tex], the lines are coincident.
Evaluating the ratios:
[tex]\[ \frac{a_1}{a_2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ \frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3} \][/tex]
[tex]\[ \frac{c_1}{c_2} = \frac{-12}{-15} = \frac{4}{5} \][/tex]
Since [tex]\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)[/tex], the lines are intersecting.
### Conclusion
The system of equations [tex]\( 4x + 3y = 12 \)[/tex] and [tex]\( 2x + 9y = 15 \)[/tex] represents intersecting lines with the solution [tex]\( x = \frac{21}{10} \)[/tex] and [tex]\( y = \frac{6}{5} \)[/tex]. Thus, the lines are intersecting.
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