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Maslowich
3 years ago
12

How do I substitute this?

Mathematics
1 answer:
Lemur [1.5K]3 years ago
7 0

You want to isolate one of the variables (x or y) so you can plug it into the other equation. The easiest one is isolating the 2nd equation.

3x² - 16x + 13 - y = 0   Add y on both sides

3x² - 16x + 13 = y

You can use this and plug it into the first equation

y - 12x + 15 = 3x²

(3x² - 16x + 13) - 12x + 15 = 3x²

3x² - 16x + 13 - 12x + 15 = 3x²  Combine like terms

3x² - 28x + 28 = 3x²  Subtract 3x² on both sides

-28x + 28 = 0  Add 28x on both sides

28 = 28x Divide 28 on both sides

1 = x

Now that you know x, you can plug it into either of the equation to find y

3(1)² - 16(1) + 13 - y = 0

3 - 16 + 13 - y = 0

-y = 0 Divide -1 on both sides

y = 0


x = 1, y = 0

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What is the solution of the system of equations?
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No solution of the system of equations y = -2x + 5 and -5y = 10x + 20 ⇒ 2nd answer

Step-by-step explanation:

Let us revise the types of solutions of a system of linear equations

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∵ y = -2x + 5

- Add 2x to both sides

∴ 2x + y = 5 ⇒ (1)

∵ -5y = 10x + 20

- Subtract 10x from both sides

∴ -10x - 5y = 20

- Divide both sides by -5

∴ 2x + y = -4 ⇒ (2)

∵ The coefficient of x in equation (1) is 2

∵ The coefficient of x in equation (2) is 2

∴ The coefficients of x in the two equations are equal

∵ The coefficient of y in equation (1) is 1

∵ The coefficient of y in equation (2) is 1

∴ The coefficients of y in the two equations are equal

∵ The numerical term in equation (1) is 5

∵ The numerical term in equation (2) is -4

∴ The numerical terms are different

From the 2nd rule above

∴ No solution of the system of equations

No solution of the system of equations y = -2x + 5 and -5y = 10x + 20

Learn more:

You can learn more about the system of equations in brainly.com/question/6075514

#LearnwithBrainly

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