Answer :
To determine the value of [tex]\( k \)[/tex] such that the system of equations
[tex]\[ 2x + 3y = 7 \][/tex]
[tex]\[ (k-1)x + (k+2)y = 3k \][/tex]
has infinitely many solutions, we need to ensure that the two equations are equivalent. This means that their coefficients must be proportional.
To find the proportionality, we consider the coefficients of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and the constant terms.
1. Coefficients of [tex]\( x \)[/tex]:
[tex]\[ \frac{2}{k-1} \][/tex]
2. Coefficients of [tex]\( y \)[/tex]:
[tex]\[ \frac{3}{k+2} \][/tex]
3. Constant terms:
The constants should follow the same ratio:
[tex]\[ \frac{7}{3k} \][/tex]
For the system to have infinitely many solutions, the ratios of the coefficients of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and the constants must all be equal. Therefore, we set up the following equality for both pairs of coefficients:
1. Equate the coefficients of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \frac{2}{k-1} = \frac{3}{k+2} \][/tex]
Cross-multiply to solve for [tex]\( k \)[/tex]:
[tex]\[ 2(k + 2) = 3(k - 1) \][/tex]
[tex]\[ 2k + 4 = 3k - 3 \][/tex]
[tex]\[ 4 + 3 = 3k - 2k \][/tex]
[tex]\[ 7 = k \][/tex]
Thus, the value of [tex]\( k \)[/tex] that makes the system have infinitely many solutions is [tex]\( k = 7 \)[/tex].
[tex]\[ 2x + 3y = 7 \][/tex]
[tex]\[ (k-1)x + (k+2)y = 3k \][/tex]
has infinitely many solutions, we need to ensure that the two equations are equivalent. This means that their coefficients must be proportional.
To find the proportionality, we consider the coefficients of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and the constant terms.
1. Coefficients of [tex]\( x \)[/tex]:
[tex]\[ \frac{2}{k-1} \][/tex]
2. Coefficients of [tex]\( y \)[/tex]:
[tex]\[ \frac{3}{k+2} \][/tex]
3. Constant terms:
The constants should follow the same ratio:
[tex]\[ \frac{7}{3k} \][/tex]
For the system to have infinitely many solutions, the ratios of the coefficients of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and the constants must all be equal. Therefore, we set up the following equality for both pairs of coefficients:
1. Equate the coefficients of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \frac{2}{k-1} = \frac{3}{k+2} \][/tex]
Cross-multiply to solve for [tex]\( k \)[/tex]:
[tex]\[ 2(k + 2) = 3(k - 1) \][/tex]
[tex]\[ 2k + 4 = 3k - 3 \][/tex]
[tex]\[ 4 + 3 = 3k - 2k \][/tex]
[tex]\[ 7 = k \][/tex]
Thus, the value of [tex]\( k \)[/tex] that makes the system have infinitely many solutions is [tex]\( k = 7 \)[/tex].
