Question #2d385

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Question #2d385

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Answer 1 · 2016-06-23T16:01:42

First part in a lot of detail to demonstrate algebraic manipulation.
Second part not so.

Point common to both equations is:
$(x, y) \to (3, 2)$ $(x,y)->(3,2)$

Explanation:

Given: $7 x + y = 23$ $" "7x+y=23$ ........................Equation (1)
$3 x = 4 y + 1$ $" "3x=4y+1$ .........................Equation (2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider equation(1)

Subtract $7 x$ $7x$ from both sides

$y = - 7 x + 23 ............................Equation (1_{a})$ $y=-7x+23" ............................Equation" (1_a)$

$Substitute for y in equation (2) using (1_{a})$ $color(green)("Substitute for "color(blue)(y)" in equation (2) using "(1_a))$

$3 x = 4 y + 1 \to 3 x = 4 (- 7 x + 23) + 1$ $color(brown)(3x=4color(blue)(y)+1" "->" "3x=4color(blue)((-7x+23))+1)$

$color(white)()$

$3 x = - 28 x + 92 + 1 \leftarrow multiplied out the bracket$ $color(brown)(3x=-28x+92+1)" "color(green)(larr" multiplied out the bracket")$
$color(white)()$

$3 x + 28 x = - 28 x + 28 x + 93 \leftarrow add 28 x to both sides$ $color(brown)(3xcolor(blue)(+28x)=-28xcolor(blue)(+28x)+93)color(green)(larr" add "color(blue)(28x)" to both sides")$
$color(white)()$

$31 x = 0 + 93$ $color(brown)(31x=0+93)$
$color(white)()$

$\frac{31}{31} x = \frac{93}{31} \leftarrow divide both sides by 31$ $color(brown)(31/(color(blue)(31)) x=93/(color(blue)(31)) )color(green)(larr" divide both sides by "color(blue)(31))$

$color(white)()$

$x = 3 \leftarrow \frac{31}{31} = 1 and \frac{93}{31} = 3$ $color(brown)(x=3)" "color(green)(larr31/31=1" and "93/31 = 3)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for $x$ $x$ in (1)

$7 x + y = 23 \Rightarrow y = 2$ $7x+y=23" " =>" " y=2$

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Question #2d385

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