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natulia [17]
3 years ago
14

What is the solution to this system of equations? x − 2y = 15 2x + 4y = -18

Mathematics
1 answer:
S_A_V [24]3 years ago
3 0
Use the subtraction method.
 2 (x - 2y = 15)      multiply top equation by 2 so we can subtract

2x - 4y = 30
2x + 4y = -18   (subtracted, we get 4x = 12...divide by 4...x = 3)

place the value for X into an original equation and solve for Y.

2(3) + 4y = -18
6 + 4y = -18
-6            -6
4y = -24  (divide by 4)

y = -6      Solution......(3, -6)
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3 years ago
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Mark's school is selling tickets to the annual dance competition. On the first day of ticket sales the school sold 4 senior tick
erica [24]

The cost of each senior ticket is $ 5 and cost of each child ticket is $ 12

<em><u>Solution:</u></em>

Let "a" be the price of each senior ticket

Let "b" be the price of each child ticket

<em><u>On the first day of ticket sales the school sold 4 senior tickets and 4 child tickets for a to total of 68</u></em>

Thus a equation is framed as:

4 senior tickets x price of each senior ticket + 4 child tickets x price of each child ticket = 68

4 \times a + 4 \times b = 68

4a + 4b = 68 ---------- eqn 1

<em><u>The school took in 120 on the second day by selling 12 senior tickets and 5 child tickets</u></em>

Similarly, we frame a equation as:

12 \times a + 5 \times b = 120

12a + 5b = 120 ---------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Multiply eqn 1 by 3</u></em>

12a + 12b = 204 -------- eqn 3

<em><u>Subtract eqn 2 from eqn 3</u></em>

12a + 12b = 204

12a + 5b = 120

( - ) --------------

7b = 84

b = 12

<em><u>Substitute b = 12 in eqn 1</u></em>

4a + 4(12) = 68

4a + 48 = 68

4a = 20

a = 5

Thus cost of each senior ticket is $ 5 and cost of each child ticket is $ 12

4 0
3 years ago
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18 + 4(28(

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18 + 112

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