solve the following system of equation a. 2x + 2y - 5 \pi = -5 x - y + \pi = 3 -3x + y + 2 \pi = -2
1. solve the following system of equation a. 2x + 2y - 5 \pi = -5 x - y + \pi = 3 -3x + y + 2 \pi = -2
1. to do manually, you can eliminate one variable from these three equations by making one variable from one equation equal to second or third equation.
and then you can repeat the process until a variable equal to a number.
2. you can use calculator.
however, there is no solution to this system of equation due to the second equation.
x - y + *?*\pi = 3
*?* means need an integer value there.
2. use the cramers rule to solve the following equation system 3x-2y=6 2x+y=11
Jawaban terlampir
Semoga membantu
3. solve simultaneously by substitution2x + 3y = 5y - 2x - 3 = 0pls bantuin cari x & y nya
2x + 3y = 5
bisa di diubah menjadi
2x = 5 - 3y
didalam bentuk kedua
y - 2x - 3 = 0
2x dapat di ubah menjadi
y - (5 - 3y) - 3 = 0
y - (-3y) - 5 - 3 = 0
4y - 8 = 0
4y = 8
y = 8/4
y = 2
y sudah di dapat
tinggal di pilih mau kerjakan bentuk pertama atau kedua
bentuk pertama
2x + 3y = 5
kita ingat y =2
2x + 3.2 = 5
2x + 6 = 5
2x = 5 - 6 = -1
x = -1/2
bentuk kedua
y - 2x - 3 = 0
2 - 2x - 3 = 0
2 - 3 - 2x = 0
-1 - 2x = 0
(-2x) = -1
x = -1/2
hasilnya tetap sama yaitu
x = -1/2 atau [tex]-\frac{1}{2}[/tex]
y = 2
4. Solve the inequalities 2x - 4< -10
2x - 4 < -10
2x < -6
x < -3
HP : {x | x < -3, x bilangan bulat}
HP : {...; -6; -5; -4}
5. Solve -8r = 64 (what is R) Solve the problem step by step.
-8 . r = 64
r = 64/-8
r = -8
Sekian terimakasih!
Semoga bermanfaat!
-8r=64
r=64/-8
r=-8
pembuktian
-8(-8)=64
-8x-8=64
64=64
6. Solve each of the following pairs of simultaneous equations. a) 2x+y= 5 x + y² = 5
Since the question is in English, i'll answer in English too.
2x+y=5 ....(1)
x+y²=5 .....(2)
modified the 2 equation
x+y²=5
x=5-y² .... (3)
subtitute (3) to (1)
2(5-y²)+y=5
10-2y²+y=5
-2y²+y+5=0
2y²-y-5=0
using Quadratic Formula, we get the roots of the quadratic equation are :
-(-1)±√((-1)²-4.2.(-5))/2(2) =
1±√1+40/4 =
1±√41/4 = 1+√41/4 or 1-√41/4
thus, enter each of y value to (1) to get the value of x.
2x+1+√41/4=5
2x=5- (1+√41)/4
2x=19+√41/4
x=19+√41/8
same to the other roots, we get
2x+1-√41/4=5
2x=5-(1-√41)/4
x=19-√41/8
Hence, the solutions are
((19+√41)/8 ; (1+√41)/4) and ((19-√41/8);(1-√41)/4)
Happy to help <3
7. Solve the exact differential equation y^'=-(2xy^2+1)/(2x^2 y).
Materi : Persamaan Diferensial
Mungkin maksudmu
PD :
[tex]y'=-\frac{2x{y}^{2}+1}{2{x}^{2}y}[/tex]
Karena PD eksak, maka kamu harus mengubah y' menjadi dy/dx, lalu ke bentuk umum PD eksak.
[tex]\frac{dy}{dx}=-\frac{2x{y}^{2}+1}{2{x}^{2}y}\\(2{x}^{2}y)\,dy=(-(2x{y}^{2}+1))\,dx\\(2x{y}^{2}+1)\,dx+(2{x}^{2}y)\,dy=0[/tex]
Sekarang, periksa apakah PD memang eksak dengan cara :
[tex]\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}[/tex]
Misalkan, M = 2xy² + 1 dan N = 2x²y, maka :
[tex]\frac{\partial{M}}{\partial{y}}=4xy\\\frac{\partial{N}}{\partial{x}}=4xy[/tex]
Karena hasil turunannya sama, maka persamaan tersebut terdiferensial total (eksak), untuk menyelesaikan saya mulai saja dari N.
Jika [tex]\frac{\partial{u}}{\partial{y}}=N[/tex], maka [tex]u=\int{N\,dy}+l(x)[/tex]. Ini akan menghasilkan :
[tex]u=\int{(2{x}^{2}y)\,dy}+l(x)\\u={x}^{2}{y}^{2}+l(x)[/tex]
Untuk mencari konstanta l(x), kamu bisa menurunkan u terhadap x secara parsial, sehingga :
[tex]\frac{\partial{u}}{\partial{x}}=2x{y}^{2}+\frac{dl}{dx}[/tex]
Karena turunan u harus sama dengan M, maka :
[tex]2x{y}^{2}+\frac{dl}{dx}=2x{y}^{2}+1\\\frac{dl}{dx}=1\\\int{\frac{dl}{dx}}=\int{1}\\l(x)=x+c[/tex]
Substitusi konstanta l(x) yang telah didapat tadi ke u semula, sehingga :
[tex]u={x}^{2}{y}^{2}+x+c\\u-c={x}^{2}{y}^{2}+x\\{x}^{2}{y}^{2}+x=k[/tex]
Jadi, solusinya :
[tex]{x}^{2}{y}^{2}+x=k[/tex]
Semoga membantu, maaf kalau saya kurang mahir berbahasa inggris.
8. solve the 6 - 2x = 3x
Jawaban:
6 - 2x = 3x
-2x - 3x = -6
-5x = -6
x = 6/5
Penjelasan dengan langkah-langkah:
semoga membantu :)
Jawaban:
6/5
Penjelasan dengan langkah-langkah:
maaf kalo salah ya maaf
9. find the solution to the system of linear equations x+y=6 dan 2x-y=0
Jawaban:
x+y=6
2x-y=0
_____+
3x =6
x =6/3
x =2
x+y=6
2+y=6
y=6-2
y=4
hp {(x,y)}={(2,4)}
10. solve the following equationsa | 2x + 5 | = | x + 2 |
| 2x + 5 | = | x + 2 |
(2x+5)² = (x+2)²
4x² + 20x + 25 = x² + 4x + 4
4x² - x² + 20x - 4x + 25 - 4 = 0
3x² + 16x + 21 = 0
(3x+7)(x+3) = 0
3x+7 = 0
3x = -7
x = -7/3
x+3 = 0
x = -3
HP = {-3, -7/3}
(2x+5)^2 - (x+2)^2 =0
4x^2 + 20x + 25 -( x^2 +4x +4) =0
3x^2 + 16x + 21 = 0
(3x+7)(x+3) = 0 [difaktorkan]
x= -7 /3 atau x= -3
11. solve the simultaneous equations y-2x=-8 and x²-3x-y=2
[tex]y - 2x = - 8 \\ y = 2x - 8 \\ \\ {x}^{2} - 3x - y = 2 \\ {x}^{2} - 3x - (2x - 8) = 2 \\ {x}^{2} - 3x - 2x + 8 = 2 \\ {x}^{2} - 5x + 8 - 2 = 0 \\ {x}^{2} - 5x + 6 = 0 \\ (x - 3)(x - 2) = 0 \\ \\ x - 3 = 0 \\ x = 3 \\ \\ x - 2 = 0 \\ x = 2[/tex]
{2, 3}
12. find the solution to each of the following system of equations by elimination for example 2x+y=5 and 3x-2y=11
3x-2y=11 |×1
2x+y=5 |×2
⇒3x-2y =11}ini
4x+2y=10}dikurangkan ini
⇒-x=1 ⇔x=-1 dan y = 2x+y=5
2×-1+y =5
-2+y=5
y=7
13. Solve the exact differential equation y^'=-(2xy^2+1)/(2x^2 y).
Materi : Persamaan Diferensial
Mungkin maksudmu
PD :
[tex]y'=-\frac{2x{y}^{2}+1}{2{x}^{2}y}[/tex]
Karena PD eksak, maka kamu harus mengubah y' menjadi dy/dx, lalu ke bentuk umum PD eksak.
[tex]\frac{dy}{dx}=-\frac{2x{y}^{2}+1}{2{x}^{2}y}\\(2{x}^{2}y)\,dy=(-(2x{y}^{2}+1))\,dx\\(2x{y}^{2}+1)\,dx+(2{x}^{2}y)\,dy=0[/tex]
Sekarang, periksa apakah PD memang eksak dengan cara :
[tex]\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}[/tex]
Misalkan, M = 2xy² + 1 dan N = 2x²y, maka :
[tex]\frac{\partial{M}}{\partial{y}}=4xy\\\frac{\partial{N}}{\partial{x}}=4xy[/tex]
Karena hasil turunannya sama, maka persamaan tersebut terdiferensial total (eksak), untuk menyelesaikan saya mulai saja dari N.
Jika [tex]\frac{\partial{u}}{\partial{y}}=N[/tex], maka [tex]u=\int{N\,dy}+l(x)[/tex]. Ini akan menghasilkan :
[tex]u=\int{(2{x}^{2}y)\,dy}+l(x)\\u={x}^{2}{y}^{2}+l(x)[/tex]
Untuk mencari konstanta l(x), kamu bisa menurunkan u terhadap x secara parsial, sehingga :
[tex]\frac{\partial{u}}{\partial{x}}=2x{y}^{2}+\frac{dl}{dx}[/tex]
Karena turunan u harus sama dengan M, maka :
[tex]2x{y}^{2}+\frac{dl}{dx}=2x{y}^{2}+1\\\frac{dl}{dx}=1\\\int{\frac{dl}{dx}}=\int{1}\\l(x)=x+c[/tex]
Substitusi konstanta l(x) yang telah didapat tadi ke u semula, sehingga :
[tex]u={x}^{2}{y}^{2}+x+c\\u-c={x}^{2}{y}^{2}+x\\{x}^{2}{y}^{2}+x=k[/tex]
Jadi, solusinya :
[tex]{x}^{2}{y}^{2}+x=k[/tex]
Semoga membantu, maaf kalau saya kurang mahir berbahasa inggris.
14. 1. Solve the equation |x + 2| = 62. Solve the equation |3x - 2| = 2x + 43. Solve |2x - 1| = |x + 4|tolong, Kak. ini harus sekarang
Jawaban:
1. |x+2|=6
x+2=6
x= 6-2
x= 4
x+2=-6
x= -6-2
x= -8
maka hp nya adalah x= -8; x= 4
2. |3x-2|= 2x+4
3x-2= 2x+4
3x-2x= 4+2
x= 6
3x-2= -2x-4
3x+2x= -4+2
5x= -2
x= -2/5
maka hp nya adalah x= -2/5; x=6
3. |2x-1|= |x+4|
2x-1= x+4
2x-x= 4+1
x=5
2x-1= -x-4
2x+x= -4+1
3x= -3
x= -3/3
x= -1
maka hp nya adalah x= -1; x=5
15. Solve the following inequalities|3-2x|≤9 |2x-9|≥5
Jawaban:
maaf kak aku gk tw nih soalnya gkbs ktik simbolnye...
aku juga gatau ihh
Jawaban:
Ini Soal Matematika Bukan Sih ?
16. Solve the simultaneous equations 4x – 3y = 1 and 3x + y = 17by substitution !
Jawaban:
x=4
y=13
caranyaadadalamfoto
17. using the graphical method, solve each of the following pairs of simultaneous equations: a. 3x-y = 0 2x-y =1
3x-y=0
2x-y=1
Elimination. Subtitution
(3x-y = 0 - 2x-y = 0) 2x-y = 1
= x = -1 = 2(-1)-y = 1
= y = 1 - (- 2)
y = 3
so, x = -1, y = 3
18. Solve the system of equations. 2x + 3y = -1 5x - 2y = 7 A. x = -1, y = -1 B. x = 1, y = 1 C. x = -1, y = 1 D. x = 1, y = -1
Jawaban:
D
Penjelasan dengan langkah-langkah:
semoga membantu....
Jawaban:
D. x = 1, y = -1
Penjelasan dengan langkah-langkah:
2x + 3y = -1
5x - 2y = 7
eliminate
4x + 6y = -2
15x - 6y = 21
→ 19x = 19
x = 1
→ 5x - 2y = 7
5(1) - 2y = 7
5 - 2y = 7
- 2y = 7-5
- 2y = 2
y = -1
19. Solve the following equations. a. 3(10+2x) = 7(5+3x) b. y-2/4 = y+1/7 - 3/4
Solutions for the equations are:
a. x = -1/3
b. y = -1
Explanation:
The linear equations in one variable is an equation that expressed in ax + b = 0 (a and b are two integers) and x is a variable and has only one solution. For example, 3x + 5 = 11 is a linear equation having a single variable in it. Hence, this equation has only one solution, which is x is equal to two.
The standard form of linear equations in one variable is stated as:
ax + b = 0
Where, ‘a’ and ‘b’ are real numbers, and both ‘a’ and ‘b’ are not equal to zero.
SolvingLinearEquationsinOneVariable:
Solve the following equations:
a. 3(10 + 2x) = 7(5 + 3x)
Solution:
=> 30 + 6x = 35 + 21x
=> 6x - 21x = 35 - 30
=> -15x = 5
=> x = -1/3
So, the right solution for 3(10 + 2x) = 7(5 + 3x) is x = -1/3
b. (y - 2)/4 = (y + 1)/7 - 3/4
Solution:
=> Based on the equation, we know that both sides are fractions. Therefore, multiply both sides by 28.
=> 7(y - 2) = 4(y + 1) - 7(3)
=> 7y - 14 = 4y + 4 - 21
=> 7y - 4y = -17 + 14
=> 3y = -3
=> y = -1
So, the right solution for (y - 2)/4 = (y + 1)/7 - 3/4 is y = -1
Learnmore:
=>HowtoSolveLinearEquationsinOneVariable:
https://brainly.co.id/tugas/1645575https://brainly.co.id/tugas/2219874https://brainly.co.id/tugas/1599747https://brainly.co.id/tugas/23336549Hopefully can help you:)
DetailsforAnswer:
Subject:Mathematics
Class:7
QuestionCode:2
Category:Chapter 6 - Linear Equations and Inequations in One Variable
CodeofCategorization:7.2.6
Keywords:Linear Equations, Variable, Solution
#TingkatkanPrestasimu
20. Solve the system:x + 6y = 17x - 3y = 8SMP VIII
x=11
y=1
Penjelasan dengan langkah-langkah:
x + 6y = 17 .......... (1)
x - 3y = 8 ............(2)
Eliminasi kedua persamaan menghasilkan :
x + 6y = 17 .......... (1)
x - 3y = 8 ............(2)
________________ -
9 y = 9 , y = 1
Substitusi y = 1 ke Persamaan 2 menghasilkan :
x - 3 (1) = 8
x - 3 = 8
x = 11
Maka (x,y) = (11,1)
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