Question

пж помогите даю 10балов 7клас Тема:Розв'язання систем різними способами. Розв'язати систему рівнянь двома способами : способом підстановки та способом додавання 2х+7у=9, х+2у=3​

Answer 1

Ответ:

(1; 1)

Объяснение:

СПОСІБ ПІДСТАНОВКИ:

2х+7у=9

х+2у=3

• 2х+7у=9
• х=3–2у

підставимо значення х в перше рівняння:

2х+7у=9

2(3–2у)+7у=9

6–4у+7у=9

3у=9–6

3у=3

у=3÷3

у=1

підставимо значення у=1 у друге рівняння

х=3–2у=3–2•1=3–2=1

ВІДПОВІДЬ: х=1; у=1 (1; 1)

СПОСІБ ДОДАВАННЯ:

2х+7у=9

х+2у=3 | •(–2)

• 2х+7у=9
• –2х–4у= –6

складемо:

2х+(–2х)+7у+(–4у)=9+(–6)

2х–2х+7у–4у=9–6

3у=3

у=3÷3

у=1

підставимо значення у=1 у будь яке рівняння:

х+2у=3

х+2•1=3

х+2=3

х=3–2

х=1

ВІДПОВІДЬ: х=1; у=1 (1; 1)

Answer 2

Solving the System of Equations by Substitution and Addition Methods

To solve the system of equations \(2x + 7y = 9\) and \(x + 2y = 3\) using the substitution and addition methods, we can follow these steps:

Substitution Method:

1. Solve one of the equations for one variable. 2. Substitute the expression obtained in step 1 into the other equation. 3. Solve for the remaining variable. 4. Substitute the value found in step 3 into one of the original equations to find the value of the other variable.

Addition Method:

1. Multiply one or both of the equations by constants to make the coefficients of one of the variables the same (or additive inverses). 2. Add the equations together to eliminate one of the variables. 3. Solve for the remaining variable. 4. Substitute the value found in step 3 into one of the original equations to find the value of the other variable.

Let's proceed with solving the given system of equations using both methods.

Substitution Method:

First, we'll solve one of the equations for one variable. Let's solve the second equation for \(x\): \[x = 3 - 2y\]

Now, we'll substitute the expression \(3 - 2y\) for \(x\) in the first equation: \[2(3 - 2y) + 7y = 9\]

Solving for \(y\): \[6 - 4y + 7y = 9\] \[3y = 3\] \[y = 1\]

Now that we have the value of \(y\), we can substitute it back into the equation \(x = 3 - 2y\) to find the value of \(x\): \[x = 3 - 2(1)\] \[x = 3 - 2\] \[x = 1\]

So, using the substitution method, we found that \(x = 1\) and \(y = 1\).

Addition Method:

To use the addition method, we'll first multiply the second equation by 2 to make the coefficients of \(x\) the same: \[2x + 4y = 6\]

Now, we'll add the modified second equation to the first equation: \[2x + 7y + 2x + 4y = 9 + 6\] \[4x + 11y = 15\]

Solving for \(x\): \[4x = 15 - 11y\] \[x = \frac{15 - 11y}{4}\]

Substitute the expression \(\frac{15 - 11y}{4}\) for \(x\) in the original second equation: \[\frac{15 - 11y}{4} + 2y = 3\]

Solving for \(y\: \[15 - 11y + 8y = 12\] \[-3y = -3\] \[y = 1\]

Substitute the value of \(y\) back into \(x = \frac{15 - 11y}{4}\): \[x = \frac{15 - 11(1)}{4}\] \[x = \frac{15 - 11}{4}\] \[x = \frac{4}{4}\] \[x = 1\]

Using the addition method, we also found that \(x = 1\) and \(y = 1\).

Therefore, both the substitution and addition methods yield the same solution: \(x = 1\) and \(y = 1\).