# Solve the system using the substitution method. y+ 3x = 5 5x - 4y = -3

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y+ 3x = 5 .............(1)

5x - 4y = -3 .............(2)

To solve using the substitution method, we will rewrite equation (1):

y + 3x = 5

subtract 3m from both sides:

==> y= 5 - 3x

Now subsitute y value in equation (2):

==> 5x - 4y = -3

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

==> 5x - 20 + 15x = -3

==> 20x - 20 = -3

Add 20 to both sides:

==> 20x = 17

**==> x = 17/20**

==> y= (5-3x) = 5- 3*17/20

= 5- 51/20 = 49/20

**==> y= 49/20**

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**