Sagot :
3x+10y=48
7x+4y=54
multiply the first equation by 2 and the 2nd equation by 5 to remove y
6x+20y=96
35x+20y=270
29x=174
x=6
substitute to any equation
7x+4y=54
42+4y=54
4y=12
y=3
the value of x is 6 and the value of y is 3
7x+4y=54
multiply the first equation by 2 and the 2nd equation by 5 to remove y
6x+20y=96
35x+20y=270
29x=174
x=6
substitute to any equation
7x+4y=54
42+4y=54
4y=12
y=3
the value of x is 6 and the value of y is 3
Answer: x=36/7
y=9/2
Step-by-step explanation:
1 Solve for yy in 3+10y=483+10y=48.
y=\frac{9}{2}y=
2
9
2 Substitute y=\frac{9}{2}y=
2
9
into 7x+4y=547x+4y=54.
7x+18=547x+18=54
3 Solve for xx in 7x+18=547x+18=54.
x=\frac{36}{7}x=
7
36
4 Therefore,
\begin{aligned}&x=\frac{36}{7}\\&y=\frac{9}{2}\end{aligned}
x=
7
36
y=
2
9