Question

4х + 7у – 26= 0 3х – 8у +7 = 0 решите систему пж

Answer 1

Ответ:

Пошаговое объяснение:

$\left \{ {{4x+7y-26=0} \atop {3x-8y+7=0}} \right. =\left \{ {{4x=26-7y} \atop {3x-8y+7=0}} \right. =\left \{ {{x=\frac{26-7y}{4} } \atop {3x-8y+7=0}} \right.\\\\ 3(\frac{26-7y}{4} })-8y=0\\\\\frac{78-21y}{4}-8y=0\\78-21y-32y=0\\-53y=-78\\y=-78:(-53)\\y=-\frac{78}{53} \\\\x=\frac{26-(-\frac{78}{53} )}{4}\\\\ x=\frac{26+\frac{78}{53} }{4} \\\\x=\frac{\frac{1456}{53} }{4}\\\\ x=\frac{1456}{53}*\frac{1}{4} \\\\ x=\frac{364}{53} \\\\x=2\frac{46}{53}$

Answer 2

To solve the system of equations, we can use either the substitution method or the elimination method. Let's use the elimination method:

Given system of equations:

1. 4x + 7y - 26 = 0
2. 3x - 8y + 7 = 0

Step 1: Eliminate one variable by multiplying both equations to make the coefficients of one variable equal but opposite in sign.

Let's eliminate y. To do this, we need to make the coefficients of y in both equations equal but opposite in sign. We can achieve this by multiplying the first equation by 8 and the second equation by 7:

Multiply equation 1 by 8: 8(4x + 7y - 26) = 8(0) 32x + 56y - 208 = 0

Multiply equation 2 by 7: 7(3x - 8y + 7) = 7(0) 21x - 56y + 49 = 0

Step 2: Now we have two equations:

1. 32x + 56y - 208 = 0
2. 21x - 56y + 49 = 0

Step 3: Add both equations to eliminate y:

(32x + 56y - 208) + (21x - 56y + 49) = 0 + 0 32x + 21x - 208 + 49 = 0 53x - 159 = 0

Step 4: Solve for x:

53x = 159 x = 159 / 53 x = 3

Step 5: Substitute the value of x back into one of the original equations to find y. Let's use equation 1:

4x + 7y - 26 = 0 4(3) + 7y - 26 = 0 12 + 7y - 26 = 0 7y = 26 - 12 7y = 14 y = 14 / 7 y = 2

Step 6: Check the solution by substituting the values of x and y into the second equation:

3x - 8y + 7 = 0 3(3) - 8(2) + 7 = 0 9 - 16 + 7 = 0 0 = 0

Since both equations are satisfied when x = 3 and y = 2, the solution to the system of equations is x = 3 and y = 2.