Question

Solve using elimination by substitution:
$\frac{x + 7}{5} - \frac{2x - y}{4} \: = \: 3y - 5 \: and \: \frac{4x - 3}{6} + \frac{5y - 7}{2} \: = \: 18 - 5x$

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Required Answer:-

Given:

$\sf 1. \: \dfrac{x + 7}{5} - \dfrac{2x - y}{4} = 3y - 5$

$\sf 2. \: \dfrac{4x - 3}{6} + \dfrac{5y - 7}{2} = \: 18 - 5x$

To Find:

The value of x and y.

Answer:

(x, y) = (3,2)

Solution:

Given that,

$\sf \implies\dfrac{x + 7}{5} - \dfrac{2x - y}{4} = 3y - 5$

LCM of 5 and 4 is 20.,

$\sf \implies\dfrac{4(x + 7) - 5(2x - y)}{20} = 3y - 5$

Simplify.,

$\sf \implies\dfrac{4x +28- 10x + 5y}{20} = 3y - 5$

$\sf \implies\dfrac{5y - 6x + 28}{20} = 3y - 5$

$\sf \implies 5y - 6x + 28= 20(3y - 5)$

$\sf \implies 5y - 6x + 28= 60y -100$

$\sf \implies 55y + 6x-100 - 28= 0$

$\sf \implies 6x + 55y = 128 \: \bf- (i)$

Again,

$\sf \implies \dfrac{4x - 3}{6} + \dfrac{5y - 7}{2} = 18 - 5x$

LCM of 2 and 6 is 6.,

$\sf \implies \dfrac{4x - 3 + 3(5y - 7)}{6} =18 - 5x$

$\sf \implies \dfrac{4x + 15y- 24}{6} =18 - 5x$

$\sf \implies 4x + 15y- 24 =6(18 - 5x)$

$\sf \implies 4x + 15y- 24 =108 -30x$

$\sf \implies 34x + 15y =108 + 24$

$\sf \implies 34x + 15y =132 \: \bf{ - }(ii)$

Therefore,

$\sf \implies 6x + 55y = 128 \: \bf- (i)$

$\sf \implies 34x + 15y =132 \: \bf{ - }(ii)$

Now, we will solve the pair of simultaneous linear equation by elimination method,

Multiplying equation (i) by 17, we get,

$\sf \implies 17(6x + 55y) =17 \times 128$

$\sf \implies 102x + 935y=2176 \: \bf - (iii)$

Multiplying equation (ii) by -3, we get,

$\sf \implies - 3(34x + 15y) =132 \times ( - 3)$

$\sf \implies -102x - 45y = - 396 \: \bf - (iv)$

Adding equations (iii) and (iv), we get,

$\sf \implies 890y = 1780$

$\sf \implies y =2$

Substituting the value of y in (i), we get,

$\sf \implies 6x + 55 \times 2 = 128$

$\sf \implies 6x + 110 = 128$

$\sf \implies 6x = 128 - 110$

$\sf \implies 6x =18$

$\sf \implies x =3$

★ Hence, the values of x and y are 3 and 2.