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scoundrel [369]
4 years ago
10

Y=6x Y=5x+7 Substitution method

Mathematics
2 answers:
Sonja [21]4 years ago
8 0
\left \{ {{y=6x} \atop {y=5x+7}} \right. \\\\Substitution\ method\\\\
6x=5x+7\ \ \ | subtract\ 5x\\\\
x=7\\
y=6*7=42
Wewaii [24]4 years ago
5 0
\left\{\begin{array}{ccc}y=6x\\y=5x+7\end{array}\right\\\\subtitute\ y=6x\ to\ the\ second\ equation\\\\6x=5x+7\ \ \ \ \ |subtract\ 5x\ fron\ both\ sides\\\boxed{x=7}\\\\subtitute\ value\ of \x=7\ to\ the\ equation\ y=6x:\\\\y=6\cdot7\\\boxed{y=42}\\\\Answer:\\  \left\{\begin{array}{ccc}x=7\\y=42\end{array}\right
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Read 2 more answers
Given f (x ) = x^2 + 3x + 2 and g (x ) = x + 1, perform the indicated operations.
Nana76 [90]

Answer:

(i) (f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = x³ + 4·x² + 5·x + 2

Step-by-step explanation:

The given functions are;

f(x) = x² + 3·x + 2

g(x) = x + 1

(i) (f - g)(x) = f(x) - g(x)

∴ (f - g)(x) = x² + 3·x + 2 - (x + 1) = x² + 3·x + 2 - x - 1 = x² + 2·x + 1

(f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = f(x) + g(x)

∴ (f + g)(x) = x² + 3·x + 2 + (x + 1) = x² + 3·x + 2 + x + 1 = x² + 4·x + 3

(f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = f(x) × g(x)

∴ (f·g)(x) = (x² + 3·x + 2) × (x + 1) = x³ + 3·x² + 2·x + x² + 3·x + 2 = x³ + 4·x² + 5·x + 2

(f·g)(x) = x³ + 4·x² + 5·x + 2

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