2x+3y=4;(k+2)x+6y=3k+2 find the value of k if there exists unique solution
Posted by Bhavana Reddy 3 years, 6 months ago
{tex}2x + 3y - 4 = 0{/tex}
{tex}a_1= 2, b_1= 3, c_1 = 4{/tex}
{tex}(k + 2)x + 6y-(3k + 2)= 0{/tex}
{tex}a _ { 2 } = k + 2 , b _ { 2 } = 6 , c _ { 2 } = - ( 3 k + 2 ){/tex}
{tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex}
or, {tex}\frac { 2 } { k + 2 } = \frac { 3 } { 6 } = \frac { 4 } {- 3 k - 2 }{/tex}
Or, {tex}\frac { 2 } { k + 2 } = \frac { 3 } { 6 } {/tex}
or, {tex}3(k + 2) = 2 \times 6{/tex}
{tex}3k + 6 = 12{/tex}
{tex}3k = 12 - 6{/tex}
or, {tex}6 = 3k{/tex}
Hence,{tex} k = 2{/tex}
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