Question

For what value of K the equations 3x+2y = 6, (k+1)x+4y = (2k+2)
have infinite solutions​

Answer 1

Answer:

The value of k is 7.

The value of k is 7.Step-by-step explanation:

The value of k is 7.Step-by-step explanation:The given equations are

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3k

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+23

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+232(k+2)=3(k-1)2(k+2)=3(k−1)

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+232(k+2)=3(k-1)2(k+2)=3(k−1)2k+4=3k-32k+4=3k−3

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+232(k+2)=3(k-1)2(k+2)=3(k−1)2k+4=3k-32k+4=3k−3k=7k=7

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+232(k+2)=3(k-1)2(k+2)=3(k−1)2k+4=3k-32k+4=3k−3k=7k=7\frac{3}{k+2}=\frac{7}{3k}k+23=3k7

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+232(k+2)=3(k-1)2(k+2)=3(k−1)2k+4=3k-32k+4=3k−3k=7k=7\frac{3}{k+2}=\frac{7}{3k}k+23=3k73(3k)=7(k+2)3(3k)=7(k+2)

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+232(k+2)=3(k-1)2(k+2)=3(k−1)2k+4=3k-32k+4=3k−3k=7k=7\frac{3}{k+2}=\frac{7}{3k}k+23=3k73(3k)=7(k+2)3(3k)=7(k+2)9k=7k+149k=7k+14

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+232(k+2)=3(k-1)2(k+2)=3(k−1)2k+4=3k-32k+4=3k−3k=7k=7\frac{3}{k+2}=\frac{7}{3k}k+23=3k73(3k)=7(k+2)3(3k)=7(k+2)9k=7k+149k=7k+149k-7k=149k−7k=14

The value of k is 7.Step-by-step explanation:The given equations are2x+3y=72x+3y=7(k-1)x+(k+2)y=3k(k−1)x+(k+2)y=3kTwo equations have infinitely many solutions if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}a2a1=b2b1=c2c1\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}k−12=k+23=3k7\frac{2}{k-1}=\frac{3}{k+2}k−12=k+232(k+2)=3(k-1)2(k+2)=3(k−1)2k+4=3k-32k+4=3k−3k=7k=7\frac{3}{k+2}=\frac{7}{3k}k+23=3k73(3k)=7(k+2)3(3k)=7(k+2)9k=7k+149k=7k+149k-7k=149k−7k=14k=7k=7

Answer 2

Answer:

a1/a2=b1/b2

3/k+1=2/4

3/k+1=1/2

cross multiply

k+1=6

k= 6-1=5