Question

What is the solution to the system of equations? 3x + 10y = -47 5x- 7y = 40
Please and thank u

Answer 1

Answer:

x=1, y=-5

Step-by-step explanation:

Given equations are:

$3x+10y=-47\ Eqn\ 1\\and\\5x-7y=40\ Eqn\ 2$

In order to solve the equation

Multiplying Eqn 1 by 5 and eqn 2 by 3 and subtracting them

So,

Eqn 1 becomes

15x+50y=-235

Eqn 2 becomes

15x-21y=120

Subtracting 2 from a

15x+50y - (15x-21y) = -235-120

15x+50y-15x + 21y = -355

71y = -355

y = -355/71

y =-5

Putting y= -5 in eqn 1

3x+10(-5) = -47

3x -50 = -47

3x = -47+50

3x = 3

x = 3/3

x = 1

Hence the solution is:

x=1, y=-5

Answer 2

Answer: x=1; y=-5

Step-by-step explanation: To solve this system we can use differente methods, in this case we are going to use substitution:

first equation: 3x+10y=-47

second equation: 5x-7y=40

we are going to isolate x from the first equation:

3x=-47-10y

$x=\frac{-47-10y}{3}$

now we replace it in the second equation:

$5*\frac{-47-10y}{3}-7y=40$

and now we solve for y:

$\frac{-235-50y}{3} -7y=40$

$\frac{-235-50y-21y}{3}=40$

$-235-71y=40*3$

$-235-71y=120$

$-71y=120+235$

$y=-355/71$

y=-5

now we replace the value of y in the first equation and solve for x:

$x=\frac{-47-10y}{3}$

$x=\frac{-47-10(-5)}{3}$

$x=\frac{-47+50}{3}$

$x=\frac{3}{3}$

x=1