Answer:
3) [tex]x=1\\ y=5[/tex]
4) [tex]x=-2\\ y=6[/tex]
Step-by-step explanation:
3) For this system of equation you already has one variable in the second equation. Therefore, you only need to solve for [tex]y[/tex] in the second one and substitute its value into the first equation to calculate [tex]x[/tex]:
[tex]\left \{ {{7x + 3y= 22} \atop {4y=20}} \right. \\[/tex]
[tex]y=\frac{20}{4} \\ y=5[/tex]
[tex]7x + 3(5)= 22\\ 7x=7\\x=1[/tex]
4) You must add both equations to cancel the variable [tex]y[/tex] and solve for [tex]x[/tex]. Then, susbstiute the value of this variable into one of the original equations to calculate [tex]y[/tex]:
[tex]\left \{ {{-2x+y=10} \atop {4x-y=-14}} \right.[/tex]
[tex]2x=-4\\ x=-2[/tex]
[tex]-2(-2)+y=10\\ y=6[/tex]
