Depth · one concept from first principles · step 2 of 4
Write the solution in terms of the constant
The values of and that satisfy the system of equations , , are both positive. Find the range of values of the constant .
Expressions
Your answer
How to solve
Strategy Convert to Algebra — Each variable appears in exactly two of the three equations, so adding all three gives the total x+y+z at once. Subtracting each original equation from that total isolates one variable in terms of a. The positivity conditions then become two linear inequalities in a.
1STEP 1
Add all three equations
Every variable appears twice in the sum.
2(x+y+z) = 10 + 6 + a → x + y + z = 8 + a/2
The symmetric pattern makes the total easy to find.
8.EE.C.8Organize Information In More WaysAdding the three equations gives .
Why?
Each equation is a level balance, and adding the three left sides on one pan and the three equal right sides on the other keeps it level.
🧱Do the same to both sides and equality holdsA balanced scale stays level when you take the same amount off both pans or add the same amount to both.
Why?
The left side holds each letter exactly twice, which is two copies of .
Why?
The order of the terms in a sum can be rearranged freely, so the six terms can be lined up as and another .
🧱Commutativity of additionPutting a with b or b with a piles up the same total — order does not change a sum.
Why?
Two copies of the group is times that group.
🧱Distributive propertya groups of (b and c) is a groups of b plus a groups of c — you can split the group.
2STEP 2
Isolate x and y
Subtract y+z=6 from the total to get x, and subtract z+x=a to get y.
x = 8 + a/2 - 6 = 2 + a/2, y = 8 + a/2 - a = 8 - a/2
Removing the pair that leaves a variable out reveals that variable.
8.EE.C.8Convert To Algebra3STEP 3
Require x ＞ 0
Solve the inequality for a.
2 + a/2 ＞ 0 → a/2 ＞ -2 → a ＞ -4
Multiplying by the positive number 2 keeps the inequality direction.
7.EE.B.4Convert To Algebra4STEP 4
Require y ＞ 0 and combine
Solve the second inequality, then keep the values of a that satisfy both.
8 - a/2 ＞ 0 → a ＜ 16, so -4 ＜ a ＜ 16
Both conditions must hold at once, so the answer is the overlap.
7.EE.B.4Convert To AlgebraAnswer
-4 ＜ a ＜ 16
Try a=0, inside the range: x=2, y=8, z=-2, and indeed 2+8=10, 8+(-2)=6, -2+2=0. At the endpoints a=-4 gives x=0 and a=16 gives y=0, which are not positive, so the endpoints are excluded.
Takeaway
Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.
- Add all three equations
- Isolate x and y
- Require x ＞ 0
- Require y ＞ 0 and combine
Where next?
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