chanasia has 30 beads. she wants to put them in boxes, so that each box will contain the same number of beads. use factors to li
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Answer:
Explanation:
When solving systems of equations in three variables, pick two equations to work with first, solve for both variables, plug them into one of the equations you first started on, then in the end, put them altogether:
{-6<em>x</em> + 5<em>y</em> + 2<em>z</em> = -11 ←
{-2<em>x</em> + <em>y</em> + 4<em>z</em> = -9 We can eliminate the<em> </em><em>y-</em><em>values</em><em> </em>obviously
{4<em>x</em> - 5<em>y</em> + 5<em>z</em> = -4 ←
{-6<em>x</em> + 5<em>y</em> + 2<em>z</em> = -11
{4<em>x</em> - 5<em>y</em> + 5<em>z</em> = -4
___________________
-2<em>x</em> + 7<em>z</em> = -15
- 7<em>z</em> -7<em>z</em>
______________
-2<em>x</em> = -15 - 7<em>z</em>
___ ________
-2 -2
<em>x</em> = 7½ + 3½<em>z</em>
-2[7½ + 3½<em>z</em>] + <em>y</em> + 4<em>z</em> = -9
-15 - 7<em>z</em> + <em>y</em> + 4<em>z</em> = -9
-15 - 3<em>z</em> + <em>y</em> = -9
+15 + 15
_______________
-3<em>z</em> + <em>y</em> = 6
+3<em>z</em> + 3<em>z</em>
<em>y</em> = 6 + 3<em>z</em> [Plug in -1 for <em>z</em><em> </em>to give you the y-value of 3; 3 = <em>y</em>; since you now have both <em>z</em><em> </em>and <em>y</em><em> </em>terms, plug in these terms back into all three equations above to get the x-value of 4.]; 4 = <em>x</em> ⤻
-6[7½ + 3½<em>z</em>] + 5[6 + 3<em>z</em>] + 2<em>z</em> = -11
-45 - 21<em>z</em> + 30 + 15<em>z</em> + 2<em>z</em> = -11
-15 - 4<em>z</em> = -11
+15 + 15
_____________
-4<em>z</em> = 4
___ __
-4 -4
<em>z</em> = -1 [Plug this into the equation for <em>y</em>]⤻
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