Solve the system using the substitution method.

y+ 3x = 5 5x - 4y = -3

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y+3x=5 and 5x-4y=-3

We'll start with the first equation:

y+3x=5

y=5-3x (1)

Now we can substitute this value for y into the second equation:

5x-4y=-3

5x-4(5-3x)=-3

5x-20+12x=-3

17x=17

x=1

Now we can substitute back into (1) to get y:

y=5-3(1)=5-3=2

**x=1; y=2**

We'll re-write the equations of the system:

3x + y = 5

5x - 4y = -3

We can also use determinant to compute x and y.

We'll calculate the determinant of the system.

det.A = 3*(-4) - 1*5

det. A = -12-5

det.A = -17

x = det.x/det.A

det.x = 5*(-4) + 3*1

det.x = -20 + 3

det.x = -17

x = -17/-17

x = 1

y = det.y/det.A

det.y = 3*(-3) - 5*5

det. y = -9-25

det.y = -34

y = -34/-17

y = 2

The solution of the system is :{1 ; 2}.

We'll verify the solutions into equations:

3*1 + 2 = 5

5 = 5

5*1 - 4*2 = -3

-3 = -3

The system of equations y+ 3x = 5 and 5x - 4y = -3 has to be solved.

From 5x - 4y = -3 isolate x and express it in terms of y.

5x = -3 + 4y

x = (4y - 3)/5

Substitute this value of x in y + 3x = 5

y + 3*(4y - 3)/5 = 5

5y + 12y - 9 = 25

17y = 34

y = 2

x = (4*2 - 3)/5 = 1

The solution of the given system of equations is x = 1 and y = 2

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