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AnnZ [28]
3 years ago
12

Identify a possible first step using the elimination method to solve the system and then find the solution to the system.

Mathematics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:x = 1

y = 1

Step-by-step explanation:

The given system of simultaneous equations is expressed as

3x - 5y = - 2 - - - - - - - - - - - - 1

2x + y = 3 - - - - - - - - - - - - - 2

The first step is to decide on which variable to eliminate. Let us eliminate x. Then we would multiply both rows by numbers which would make the coefficients of x to be equal in both rows.

Multiplying equation 1 by 2 and equation 2 by 3, it becomes

6x - 10y = - 4

6x + 3y = 9

Subtracting, it becomes

- 13y = - 13

y = - 13/- 13 = 1

The next step is to substitute y = 1 into any of the equations to determine x.

Substituting y = 1 into equation 2, it becomes

2x + 1 = 3

2x = 3 - 1 = 2

x = 2/2 = 1

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x(3x-1)

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take the similar ones out

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3 years ago
(−4 + 5) − (6 + 7) = x x 0
weeeeeb [17]

1

Add the numbers

(

−

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−

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({\color{#c92786}{-4}}+{\color{#c92786}{5}})-(6+7)=xx^{0}

(−4+5)−(6+7)=xx0

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({\color{#c92786}{1}})-(6+7)=xx^{0}

(1)−(6+7)=xx0

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Add the numbers

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−

(

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+

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1-({\color{#c92786}{6}}+{\color{#c92786}{7}})=xx^{0}

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1-({\color{#c92786}{13}})=xx^{0}

1−(13)=xx0

3

Multiply the numbers

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1

⋅

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1{\color{#c92786}{-1}} \cdot {\color{#c92786}{13}}=xx^{0}

1−1⋅13=xx0

1

−

1

3

=

0

1{\color{#c92786}{-13}}=xx^{0}

1−13=xx0

4

Subtract the numbers

1

−

1

3

=

0

{\color{#c92786}{1-13}}=xx^{0}

1−13=xx0

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1

2

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{\color{#c92786}{-12}}=xx^{0}

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Combine exponents

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1

2

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-12={\color{#c92786}{xx^{0}}}

−12=xx0

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2

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-12={\color{#c92786}{x^{1}}}

−12=x1

Show less

Solution

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2

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1

3 0
3 years ago
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