Question

Solve simultaneous equation 4x+3y-4=0;6x=8-5y


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Answer 1

Explanation:

Given, 4x + 3y - 4 = 0 -------(1)

6x = 8 - 5y

6x + 5y - 8 = 0 -------(2)

use Cramer's rule ,

4x + 3y - 4 = 0

6x + 5y - 8 = 0

for equations ,

\frac{x}{3\times-8-(-4)\times5}=\frac{-y}{4\times - 8-(-4)\times6}=\frac{1}{4\times5-3\times6}

=> x/(-24 + 20) = -y/(-32+ 24) = 1/(20 - 18)

=> x/-4 = -y/-8= 1/2

=> x/-4 = y/8 = 1/2

=> x = -4/2 = -2 and y = 8/2 = 4

hence, x = -2 and y = 4

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\: \: \: \: \: \: \: \: \bold{4x + 3y-4=0} \\ \\ </p><p> \bold {=> \underline{ \red{ \: \: 4x+3y = 4 \: \: }}\: \: \: \: ----->(1)} \\ \\ \\ \: \: \: \: \: \: \: \bf{</p><p>6x=8-5y } \\ \\ \bold{</p><p>=> \underline{ \red{ \: \: 6x+5y = 8 \: \: }}\: \: \: \: ----->(2)}$

$\bold{(1) \times \red{5} => 20x+15y=20 \: \: \: ----->(3)} \\ \\ </p><p> \bold{(2) \times \red{3}=> 18x+15y=24 \: \: \: ----->(4)}$

Now,

$\bold{(3)-(4)=> 2x = - 4} \\ \\ \bold{</p><p>=> x = \: \frac{ - 4}{2} } \\ \\ \bold{ \therefore \: \underline{ \red{ \: \: x = -2 \: \: }}}</p><p>$

Putting the value of x in eq. (1)

$\bold{4 \times (-2)+3y=4} \\ \\ \bold{</p><p>=> -8 +3y=4} \\ \\ \bold{</p><p>=>3y=4+8} \\ \\ \bold{</p><p>=>3y=12} \\ \\ \bold{</p><p>=> y = \frac{12}{3} } \\ \\ \bold{</p><p> \therefore \: \underline{ \red{ \: \: y=4 \: \: }}}</p><p>$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: VERIFICATION \: \: } \mid}}}}}$

We get,

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{\red{ \: \: \: x = -2 \: , \: y= 4 \: \: }}}</p><p>$

Putting the value of x and y in eq. (1)

$\mathsf{L.H.S. = 4 \times( \red{-2})+3 \times( \red{4})} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \mathsf{</p><p>= -8 + 12} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \mathsf{</p><p>= 4 \: =R.H.S.}</p><p>$

Again,

Putting the value of x and y in eq. (2)

$\mathsf{L.H.S. = 6 \times ( \red{-2})+5 \times ( \red{4})} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \mathsf{</p><p>= -12+20} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \mathsf{</p><p>=8=R.H.S.}</p><p>$

As in both the equation, L.H.S.=R.H.S.

So,

$\bold{ Value \: \: \: of \: \: \underline{\pink{ \: \: x = -2 \: \: }} \: \: \: and \: \: \: Value \: \: \: of \: \: \underline{ \pink{ \: \: y = 4 \: \: }}}$