Depth · one concept from first principles · step 2 of 4
Solve in terms of a, then demand positivity
For the system , the solution values , , are all positive. How many integers make this possible?
Expressions
Your answer
How to solve
Strategy Convert to Algebra — The system is symmetric: each variable appears in exactly two equations. Adding all three gives x+y+z, and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in a.
1STEP 1
Add all three equations
Each of x, y, z appears twice on the left.
2(x+y+z)=6+10+a=16+a → x+y+z=8+a/2
The total of all three is the easiest thing to find first.
8.EE.C.8Convert To AlgebraAdding the three equations gives .
Why?
Each left side equals its right side, and adding equal amounts to equal amounts keeps the two totals equal.
🧱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?
Each of , , appears in exactly two of the three equations, so the added left sides hold two of each, which is twice .
Why?
The six terms can be reordered and regrouped into two copies of without changing the total.
🧱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 , two of and two of together are two groups of .
🧱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 each variable
Subtract each original equation from the total.
x=(8+a/2)-10=a/2-2, y=(8+a/2)-a=8-a/2, z=(8+a/2)-6=2+a/2
Removing the pair that leaves out a variable leaves exactly that variable.
8.EE.C.8Identify Subproblems3STEP 3
Require positivity
Set each expression greater than 0 and solve.
a/2-2 ＞ 0 → a ＞ 4, 8-a/2 ＞ 0 → a ＜ 16, 2+a/2 ＞ 0 → a ＞ -4
All three must hold, so the binding conditions are a ＞ 4 and a ＜ 16: 4 ＜ a ＜ 16.
7.EE.B.4Convert To Algebra4STEP 4
Count the integers
The integers with 4 ＜ a ＜ 16 are 5, 6, …, 15.
15-5+1=11
Both ends are excluded, so 4 and 16 are not counted.
7.EE.B.4Make A Systematic ListAnswer
11
Check the extremes: a=5 gives x=1/2, y=11/2, z=9/2, all positive; a=4 gives x=0 (not positive), and a=16 gives y=0 (not positive). So 5 to 15 is exactly right.
Takeaway
First let the letter a ride along as if it were a number, then ask what it must be.
- Add all three equations
- Isolate each variable
- Require positivity
- Count the integers
Where next?
Level upsuggested
Another one like thissuggested
Step back and solidifysuggested
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