Question

3y-4+x=0 5x+6y=11 ????

Answer 1

Answer:

(1, 1)

Step-by-step explanation:

Given system of linear equations:

$\begin{cases}3y-4+x=0\\5x+6y=11\end{cases}$

Rewrite the first equation to make x the subject:

$\implies 3y-4+x=0$

$\implies 3y-4+x+4=0+4$

$\implies 3y+x=4$

$\implies 3y+x-3y=4=3y$

$\implies x=4-3y$

Substitute the expression for x into the second equation and solve for y:

$\implies 5(4-3y)+6y=11$

$\implies 20-15y+6y=11$

$\implies 20-9y=11$

$\implies 20-9y-20=11-20$

$\implies -9y=-9$

$\implies -9y \div -9=-9 \div -9$

$\implies y=1$

Substitute the found value of y into the rearranged first equation and solve of x:

$\implies x=4-3(1)$

$\implies x=4-3$

$\implies x=1$

Therefore, the solution to the system of equations is (1, 1).

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Answer 2

Answer:

x = 1

y = 1

Step-by-step explanation:

3y - 4 + x = 0

x = -3y + 4

5x + 6y = 11

5(-3y + 4) + 6y = 11

-15y + 20 + 6y = 11

-9y = -9

y = 1

3(1) - 4 + x = 0

x = 4 - 3

x = 1