Simultaneous equations , could you show me how to work it out? #5x+2y=7# #4x-3y=24#

There are two main ways to solve an equation system. The first is substitution which works for nearly all equation systems, but is more tedious, and then you can also add or subtract the equations from each other (since both sides are equal).

In this case I can see that we can subtract the equations to cancel out for #y#, but we need to multiply the equations by #3# and #2#:

#Eq_1:# #3(5x+2y)=7*3#

#Eq_2:# #2(4x-3y)=24*2#

#Eq_1:# #15x+6y=21#

#Eq_2:# #8x-6y=48#

Now I see that the #y#'s will cancel if I add the two equations, so I'll do just that:
#15x+cancel(6y)+8x-cancel(6y)=21+48#

#15x+8x=69#

#23x=69#

#x=69/23#

#x=3#

And then we can just plug in for #x# in one of the equations and solve for #y#:

#Eq_2:# #5*3+2y=7#

#15+2y=7#

#2y=7-15#

#2y=-8#

#y=-4#

#"one approach is the "color(blue)"elimination method"#

#5x+2y=7to(1)#

#4x-3y=24to(2)#

#"to eliminate the y-term we require their coefficients to"#
#"have the same numeric value but with different signs"#

#"multiply "(1)" by 3 and "(2)" by 2"#

#15x+6y=21to(3)#

#8x-6y=48to(4)#

#"add "(3)" and "(4)" term by term to eliminate y"#

#(15x+8x)+(6y-6y)=(21+48)#

#rArr23x=69#

#"divide both sides by 23"#

#(cancel(23) x)/cancel(23)=69/23#

#rArrx=3#

#"substitute this value into either "(1)" or "(2)#

#(1)to15+2y=7#

#rArr2y=7-15=-8#

#rArry=-4#

#"the point of intersection of the 2 lines "=(3,-4)#
graph{(y+5/2x-7/2)(y-4/3x+8)((x-3)^2+(y+4)^2-0.04)=0 [-10, 10, -5, 5]}