Whats New - June 2018 - Free Printable

Explanation:
Let’s solve each part one by one using elimination — the method where we add or subtract equations to cancel out one variable.

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1a)
Equations:
① $4a + 3b = 44$
② $3a + 3b = 39$

Subtract equation ② from ①:
$(4a - 3a) + (3b - 3b) = 44 - 39$
→ $a = 5$

Now plug $a = 5$ into equation ②:
$3(5) + 3b = 39$
→ $15 + 3b = 39$
→ $3b = 24$
→ $b = 8$

✔ Check in ①: $4(5) + 3(8) = 20 + 24 = 44$ ✔️

Answer for 1a: $a = 5$, $b = 8$

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1b)
Equations:
① $2a + b = 23$
② $a + b = 16$

Subtract ② from ①:
$(2a - a) + (b - b) = 23 - 16$
→ $a = 7$

Plug into ②:
$7 + b = 16$ → $b = 9$

✔ Check in ①: $2(7) + 9 = 14 + 9 = 23$ ✔️

Answer for 1b: $a = 7$, $b = 9$

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1c)
Equations:
① $5a + 2b = 27$
② $5a + 4b = 49$

Subtract ① from ②:
$(5a - 5a) + (4b - 2b) = 49 - 27$
→ $2b = 22$ → $b = 11$

Plug into ①:
$5a + 2(11) = 27$ → $5a + 22 = 27$ → $5a = 5$ → $a = 1$

✔ Check in ②: $5(1) + 4(11) = 5 + 44 = 49$ ✔️

Answer for 1c: $a = 1$, $b = 11$

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1d)
Equations:
① $3a + 2b = 16$
② $4a + 6b = 28$

Let’s eliminate $b$. Multiply ① by 3:
→ $9a + 6b = 48$ (call this ③)

Now subtract ② from ③:
$(9a - 4a) + (6b - 6b) = 48 - 28$
→ $5a = 20$ → $a = 4$

Plug into ①:
$3(4) + 2b = 16$ → $12 + 2b = 16$ → $2b = 4$ → $b = 2$

✔ Check in ②: $4(4) + 6(2) = 16 + 12 = 28$ ✔️

Answer for 1d: $a = 4$, $b = 2$

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1e)
Equations:
① $2a + b = 14$
② $4a + 6b = 36$

Eliminate $a$: Multiply ① by 2 → $4a + 2b = 28$ (call this ③)

Subtract ③ from ②:
$(4a - 4a) + (6b - 2b) = 36 - 28$
→ $4b = 8$ → $b = 2$

Plug into ①:
$2a + 2 = 14$ → $2a = 12$ → $a = 6$

✔ Check in ②: $4(6) + 6(2) = 24 + 12 = 36$ ✔️

Answer for 1e: $a = 6$, $b = 2$

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1f)
Equations:
① $6a + 5b = 39$
② $2a + 3b = 17$

Let’s eliminate $a$. Multiply ② by 3:
→ $6a + 9b = 51$ (call this ③)

Subtract ① from ③:
$(6a - 6a) + (9b - 5b) = 51 - 39$
→ $4b = 12$ → $b = 3$

Plug into ②:
$2a + 3(3) = 17$ → $2a + 9 = 17$ → $2a = 8$ → $a = 4$

✔ Check in ①: $6(4) + 5(3) = 24 + 15 = 39$ ✔️

Answer for 1f: $a = 4$, $b = 3$

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2. Why can’t we solve:
$$
\begin{cases}
x + y = 9 \\
2x + 2y = 18
\end{cases}
$$

Let’s try elimination: Multiply first equation by 2:
→ $2x + 2y = 18$

Now compare with second equation: it’s exactly the same!
So both equations represent the same line — infinitely many solutions (any point on that line works), not a unique solution.

In other words: the two equations are dependent — one is just a multiple of the other. So we can’t find *one* answer for $x$ and $y$; there are infinitely many pairs $(x, y)$ that satisfy both (e.g., $x=0, y=9$; $x=5, y=4$; etc.).

That’s why it’s “not possible” to solve for a single unique solution.

Final Answer:
1a) $a = 5,\ b = 8$
1b) $a = 7,\ b = 9$
1c) $a = 1,\ b = 11$
1d) $a = 4,\ b = 2$
1e) $a = 6,\ b = 2$
1f) $a = 4,\ b = 3$
2) The two equations are multiples of each other (dependent), so they represent the same line — infinitely many solutions, not a unique one.