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Sagot :
To determine the value of [tex]\( k \)[/tex] for which the system of linear equations
[tex]\[ 6x - 2y - 3 = 0 \][/tex]
[tex]\[ kx - y - 2 = 0 \][/tex]
has a unique solution, follow these steps:
1. Rewrite the equations in the standard form [tex]\( ax + by = c \)[/tex]:
[tex]\[ 6x - 2y = 3 \][/tex]
[tex]\[ kx - y = 2 \][/tex]
2. Identify the coefficients:
For the first equation [tex]\( 6x - 2y = 3 \)[/tex]:
- [tex]\(a_1 = 6\)[/tex]
- [tex]\(b_1 = -2\)[/tex]
- [tex]\(c_1 = 3\)[/tex]
For the second equation [tex]\( kx - y = 2 \)[/tex]:
- [tex]\(a_2 = k\)[/tex]
- [tex]\(b_2 = -1\)[/tex]
- [tex]\(c_2 = 2\)[/tex]
3. Condition for a unique solution:
A pair of linear equations will have a unique solution if the ratios of the coefficients [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are not equal, i.e.,
[tex]\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \][/tex]
4. Set up the inequality:
[tex]\[ \frac{6}{k} \neq \frac{-2}{-1} \][/tex]
5. Simplify the inequality:
[tex]\[ \frac{6}{k} \neq 2 \][/tex]
6. Solve for [tex]\( k \)[/tex]:
[tex]\[ 6 \neq 2k \][/tex]
Divide both sides by 2:
[tex]\[ 3 \neq k \][/tex]
Thus, the value of [tex]\( k \)[/tex] for which the system of linear equations will have a unique solution is:
[tex]\[ k \neq 3 \][/tex]
Therefore, the value that [tex]\( k \)[/tex] cannot equal for the equations to have a unique solution is [tex]\( \boxed{3} \)[/tex].
[tex]\[ 6x - 2y - 3 = 0 \][/tex]
[tex]\[ kx - y - 2 = 0 \][/tex]
has a unique solution, follow these steps:
1. Rewrite the equations in the standard form [tex]\( ax + by = c \)[/tex]:
[tex]\[ 6x - 2y = 3 \][/tex]
[tex]\[ kx - y = 2 \][/tex]
2. Identify the coefficients:
For the first equation [tex]\( 6x - 2y = 3 \)[/tex]:
- [tex]\(a_1 = 6\)[/tex]
- [tex]\(b_1 = -2\)[/tex]
- [tex]\(c_1 = 3\)[/tex]
For the second equation [tex]\( kx - y = 2 \)[/tex]:
- [tex]\(a_2 = k\)[/tex]
- [tex]\(b_2 = -1\)[/tex]
- [tex]\(c_2 = 2\)[/tex]
3. Condition for a unique solution:
A pair of linear equations will have a unique solution if the ratios of the coefficients [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are not equal, i.e.,
[tex]\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \][/tex]
4. Set up the inequality:
[tex]\[ \frac{6}{k} \neq \frac{-2}{-1} \][/tex]
5. Simplify the inequality:
[tex]\[ \frac{6}{k} \neq 2 \][/tex]
6. Solve for [tex]\( k \)[/tex]:
[tex]\[ 6 \neq 2k \][/tex]
Divide both sides by 2:
[tex]\[ 3 \neq k \][/tex]
Thus, the value of [tex]\( k \)[/tex] for which the system of linear equations will have a unique solution is:
[tex]\[ k \neq 3 \][/tex]
Therefore, the value that [tex]\( k \)[/tex] cannot equal for the equations to have a unique solution is [tex]\( \boxed{3} \)[/tex].
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