solve the system of equations by substitution. 3/8 x + 1/3 y =17/24 and x + 7y = 8

[tex] \left \{ {{ \frac{3}{8}x+ \frac{1}{3}y = \frac{17}{14} } \atop {x+7y=8}} \right. [/tex]

To solve this system by substitution, first isolate x in the second equation.

[tex]x+7y=8[/tex]
[tex]x+7y-7y=8-7y[/tex]
[tex]x=8-7y[/tex]

Now, plug this expression ([tex]8-7y[/tex]) for x in the top equation to solve for y.

[tex] \frac{3}{8} (8-7y)+ \frac{1}{3} y= \frac{17}{14} [/tex]
[tex]3- \frac{21}{8} y+ \frac{1}{3} y= \frac{17}{24} [/tex]
[tex]72-63y+8y=17[/tex]
[tex]72-55y=17[/tex]
[tex]-55y=17-72[/tex]
[tex]-55y=-55[/tex]
[tex]y=1[/tex]

Now that you have y, plug it into the second equation and solve for x.

[tex]x+7y=8[/tex]
[tex]x+7(1)=8[/tex]
[tex]x+7=8[/tex]
[tex]x=1[/tex]

Last step is to plug your x- and y-values in to both equations to check your work.

[tex] \frac{3}{8} (1)+ \frac{1}{3} y= \frac{17}{24} [/tex]

[tex] \frac{3}{8} * \frac{3}{3} = \frac{9}{24} ; \frac{1}{3} * \frac{8}{8} = \frac{8}{24} [/tex]

[tex] \frac{9}{24} + \frac{8}{24} = \frac{17}{24} [/tex]  <--True

[tex]1+7(1)=8[/tex]
[tex]1+7=8[/tex]   <--True

Answer:
[tex]x=1 \\ y=1[/tex]