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miskamm [114]
4 years ago
13

Find the illegal values of c in the multiplication statement c^2-3c-10/c^2+5c-14-c^2-c-2/c^2-2c-15

Mathematics
1 answer:
matrenka [14]4 years ago
3 0

<span>c^2 + 5c - 14 = 0 </span>
<span>(c-2)(c+7) = 0 </span>
<span>c-2 = 0 or c+7 = 0 </span>
<span>c = 2 or c = -7 </span>

<span>c^2 - 2c - 15 = 0 </span>
<span>(c-5)(c+3) = 0 </span>
<span>c-5 = 0 or c+3 = 0 </span>
<span>c = 5 or c = -3</span>
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I will try to solve your system of equations.
3
x
−
4
y
=
2
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x
−
12
y
=
6
Step: Solve
3
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−
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y
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y
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4
y
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+
4
y
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y
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2
3
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y
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3
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3
y
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3
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x
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9
x
−
12
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6
:
9
x
−
12
y
=
6
9
(
4
3
y
+
2
3
)
−
12
y
=
6
6
=
6
(Simplify both sides of the equation)
6
+
−
6
=
6
+
−
6
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