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Snezhnost [94]
3 years ago
12

Clip 102

Mathematics
1 answer:
PtichkaEL [24]3 years ago
8 0

Answer:

Let's solve your system by elimination.

2a + 3c =66 ; 3a + 5c = 105

Multiply the first equation by 5,and multiply the second equation by -3.

5 (2a + 3c = 66)

−3 (3a + 5c = 105)

Becomes:

10a + 15c = 330

−9a − 15c = −315

Add these equations to eliminate c:

a = 15

Now that we've found a let's plug it back in to solve for c.

Write down an original equation:

2a + 3c = 66

Substitute 15 for a in 2a + 3c = 66:

(2)(15) + 3c = 66

3c + 30 = 66 (Simplify both sides of the equation)

3c + 30 −30 = 66 −30 (Add -30 to both sides)

3c = 36

3c3= 363 (Divide both sides by 3)

c = 12

Answer:

a = 15 and c = 12

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