Question

{x+4y/9-x-2y/3=0 {x+7y/4+x+y/6=4 ОТВЕТ ДОЛЖЕН БЫТЬ (5,1) Смотреть ответ Участник Знаний Участник Знаний X+4y:9-x-2y:3=0 // умножаем обе части на 9 9х+4у-9х-6у=0 -2у=0 у=0 х+7у:4+х+у:6=4 // умножаем обе части на 12 12х+21у+12х+2у=48 24х+23у=48 так как у =0 24х=48 х=2 ответ (2;0) может конечно неправильно

Answer 1

To verify the given system of equations and find the solution, let's re-evaluate the equations step by step:

The system of equations is given as:

1. $\frac{x+4y}{9} - \frac{x-2y}{3} = 0$
2. $\frac{x+7y}{4} + x + \frac{y}{6} = 4$

Now, we'll solve it step by step:

1. $\frac{x+4y}{9} - \frac{x-2y}{3} = 0$

To simplify the equation, let's get rid of the fractions by multiplying both sides by 9 (the least common multiple of 9 and 3): $9 \left( \frac{x+4y}{9} \right) - 9 \left( \frac{x-2y}{3} \right) = 0$

Simplifying further: $x + 4y - 3x + 6y = 0$ $-2x + 10y = 0$ $-2x = -10y$ $x = 5y$ .......... (Equation 1)

2. $\frac{x+7y}{4} + x + \frac{y}{6} = 4$

Similarly, let's eliminate the fractions by multiplying both sides by 12 (the least common multiple of 4 and 6): $12 \left( \frac{x+7y}{4} \right) + 12x + 2y = 48$

Simplifying further: $3x + 21y + 12x + 2y = 48$ $15x + 23y = 48$ ........... (Equation 2)

Now, let's substitute the value of $x$ from Equation 1 into Equation 2: $15(5y) + 23y = 48$ $75y + 23y = 48$ $98y = 48$ $y = \frac{48}{98} \approx 0.49$

Now, we can find the value of $x$ using Equation 1: $x = 5y$ $x = 5 \cdot 0.49 \approx 2.45$

So, the solution to the system of equations is approximately (2.45, 0.49), not (5, 1) as given in the initial answer.

It looks like the initial answer provided was incorrect. The correct solution is approximately (2.45, 0.49).