How do you solve the system of equations $10 x - 9 y = - 3$ $10x - 9y = - 3$ and $- 2 x + 3 y = 9$ $- 2x + 3y = 9$ ?

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How do you solve the system of equations $10 x - 9 y = - 3$ $10x - 9y = - 3$ and $- 2 x + 3 y = 9$ $- 2x + 3y = 9$ ?

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Answer 1 · 2016-11-06T11:31:09

$x = 6 and y = 7$ $x=6 and y=7$

Explanation:

Solving this system is determined by removing or eliminating one of the unknowns and finding the second.

After finding the value of one unknown we substitute its value to find the value of the variable we removed first.

$10 x - 9 y = - 3$ $10x - 9y =-3$ $E Q 1$ $" "EQ1$

$- 2 x + 3 y = 9$ $-2x + 3y =9$ $E Q 2$ $" " EQ2$

Method:
$" "$

First:
Multiply $EQ2$ $" EQ2 "$ by $3$ $color(red)" 3 "$

The coefficient of $y$ $" y "$ in $EQ2$ $" EQ2 "$ will be opposite to that in $EQ1$ $" EQ1 "$

Multiplying $EQ2$ $" EQ2 "$ by $3$ $color(red)" 3 "$ gives:

$- 6 x + 9 y = 27 E Q 2$ $-6x+9y=27" "EQ2"$

$10 x - 9 y = - 3 EQ1$ $10x - 9y =-3" EQ1 "$
$" "$

Second:
Add both equations

$EQ2 + E Q 1$ $"EQ2" + EQ1"$

$- 6 x + 10 x + 9 y - 9 y = 27 - 3$ $-6x +10x+9y-9y=27-3$

$\Rightarrow 4 x + 0 y = 24$ $rArr4x+0y = 24$

$\Rightarrow 4 x = 24$ $rArr 4x = 24$

$\Rightarrow x = \frac{24}{4}$ $rArr x=24/4$

$\Rightarrow x = 6$ $rArr color(blue)(x=6)$
$" "$

Third:
Substitute $x = 6$ $x=6$ in $EQ1$ $" EQ1 "$ to find $y$ $y$

$10 x - 9 y = - 3 EQ1$ $10x - 9y =-3" EQ1 "$

$\Rightarrow 10 (6) - 9 y = - 3$ $rArr10(6)-9y=-3$

$\Rightarrow 60 - 9 y = - 3$ $rArr60-9y=-3$

$\Rightarrow - 9 y = - 3 - 60$ $rArr-9y=-3-60$

$\Rightarrow - 9 y = - 63$ $rArr-9y=-63$

$\Rightarrow y = \frac{- 63}{- 9}$ $rArry=(-63)/-9$

$\Rightarrow y = 7$ $rArrcolor(blue)(y=7)$

Therefore,

$x = 6 and y = 7$ $x=6 and y=7$

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How do you solve the system of equations $10 x - 9 y = - 3$ $10x - 9y = - 3$ and $- 2 x + 3 y = 9$ $- 2x + 3y = 9$ ?

1 Answer

Explanation:

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How do you solve the system of equations $10 x - 9 y = - 3$ $10x - 9y = - 3$ and $- 2 x + 3 y = 9$ $- 2x + 3y = 9$ ?