The answer is: [D]: " (-3, 5) " .
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We are given the following two equations:
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x + y = 2 ;
y = -2x - 1 ;
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Let us rewrite the first equation:
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x + y = 2 ; solve in terms of "y" ;
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Subtract "x" from EACH SIDE of the equation; to isolate "y" on one side of the equation ; and to solve in terms of "y" ;
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x + y - x = 2 - x ;
y = 2 - x ;
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Now, let us rewrite this equation, and the second equation:
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y = 2 - x ;
y = - 2x - 1
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Start with "Choice [A]: (5, -3) " ; When x = 5, y = 3. Do these values hold true for BOTH of the two (2) equations?
Start with: " y = 2 - x " ; "-3 =? 2 - 5 ? " ; "-3 =? -7 ? " No! So we can rule out "Choice [A]" without even considering the second equation.
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Then, consider: "Choice [B]: (-1, 3) " ; When x = -1, y = 3. Do these values hold true for BOTH of the two (2) equations?
Start with " y = 2 - x " . "3 =? 2- (-1) ? " ; "3 =? 2 + 1 ? Yes!
Now, let us try plugging these values; "x = -1" and "y = 3" into the second equation:
" y = - 2x - 1 " ; "3 =? -2(-1) - 1 ? " ; "3 =? 2 - 1 ? " ; "3 =? 1 ? No!
So we can rule out "Choice [B]"
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Then, consider: "Choice [C]: (3, -1) " ; When x = 3, y = -1. Do these values hold true for BOTH of the two (2) equations?
Start with " y = 2 - x " . "-1 =? 2 - (3) ? " ; "-1 =? -1 ? Yes!
Now, let us try plugging these values; "x = 3" and "y = -1" into the second equation:
" y = - 2x - 1 " ; "-1 =? -2(3) - 1 ? " ; "-1 =? -6 - 1 ? " ; "-1 =? -7 ? No!
So we can rule out "Choice [C]" .
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This leaves us with: "Choice [D]". But let us try working out this answer choice.
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Consider "Choice [D]: (-3, 5) " ; When x = -3, y = 5. Do these values hold true for BOTH of the two (2) equations?
Start with " y = 2 - x " . "5 =? 2 - (-3) ? " ; "5 =? 2 + 3 ? " ; "5 =? 5 ?" ; Yes!
Now, let us try plugging these values; "x = 3" and "y = -1" into the second equation:
" y = - 2x - 1 " ; "5 =? -2(-3) - 1 ? " ; "5 =? 6 - 1 ? " ; "5 =? 5" ? ; Yes!
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So, Answer choice: [D]: " (-3, 5) " — is the correct answer.
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