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Sagot :
[tex]\left \{ {3x-4y=100 \atop \frac{x}3+\frac{4y}3=100[/tex]
Let's use substitution to solve.
First, get one of the variables on its own side in both equations.
[tex]\left \{ {3x=100+4y \atop \frac{x}3=100-\frac{4y}3[/tex]
Now, we want both of the equations to have the same value of x.
Let's multiply the bottom equation by 9 to accomplish this.
[tex]\left \{ {3x=100+4y \atop 3x=900-12y[/tex]
Now we can set the right side of both equations equal to each other (substitution)
[tex]100+4y=900-12y[/tex]
Let's solve for y.
[tex]100+16y=900 \\ 16y=800 \\ \boxed{y=50}[/tex]
Now we can use [tex]y=25[/tex] in an earlier equation to find x.
[tex]3x-4y=100 \\ 3x-4*50=100 \\ 3x-200 = 100 \\ 3x=300 \\ \boxed{x=100}[/tex]
Let's use substitution to solve.
First, get one of the variables on its own side in both equations.
[tex]\left \{ {3x=100+4y \atop \frac{x}3=100-\frac{4y}3[/tex]
Now, we want both of the equations to have the same value of x.
Let's multiply the bottom equation by 9 to accomplish this.
[tex]\left \{ {3x=100+4y \atop 3x=900-12y[/tex]
Now we can set the right side of both equations equal to each other (substitution)
[tex]100+4y=900-12y[/tex]
Let's solve for y.
[tex]100+16y=900 \\ 16y=800 \\ \boxed{y=50}[/tex]
Now we can use [tex]y=25[/tex] in an earlier equation to find x.
[tex]3x-4y=100 \\ 3x-4*50=100 \\ 3x-200 = 100 \\ 3x=300 \\ \boxed{x=100}[/tex]
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