How do you solve this 🤔?
Solve these simultaneous equations:
A. 7c + 3d = 29
5c - 4d = 33
And
B . 2x +3y = 11
3x - 5y = -12
Solve these simultaneous equations:
A. 7c + 3d = 29
5c - 4d = 33
And
B . 2x +3y = 11
3x - 5y = -12
2 Answers
Simple simultaneous equations are solved by using algebra to equate the coefficients of one of the variables. Then one equation is subtracted from the other to find one of the variables.
Explanation:
7c + 3d = 29
5c - 4d = 33
Multiplying Eq1 by 4 and Eq2 by 3 will make the 'd' coefficients equal.
Because Eq2 'd' coefficient is already negative, we can just ADD the two equations to obtain:
Substitute this back into Eq2 to find d:
ALWAYS CHECK the result by putting both answers back into Eq1:
Now, you can try it with the second set.
And
B . 2x +3y = 11
3x - 5y = -12
Explanation:
#"there are 2 methods for solving systems of equations"#
#"substitution or elimination"#
#"the questions here are both suited for elimination"#
#(A)#
#7c+3d=29to(1)#
#5c-4d=33to(2)#
#"since the coefficients of d have opposing signs if we"#
#"multiply by suitable values we can eliminate d"#
#"multiply "(1)" by 4 and "(2)" by 3"#
#rArr28c+12d=116to(3)#
#rArr15c-12d=99to(4)#
#"note the coefficients of d are + 12 and - 12, so"#
#"adding them will eliminate the term in d"#
#"adding "(3)" and "(4)" term by term gives"#
#(28c+15c)+(+12d-12d)=(116+99)#
#rArr43c=215#
#"divide both sides by 43"#
#(cancel(43) c)/cancel(43)=215/43#
#rArrc=5#
#"to obtain the value for d substitute this value into"#
#"either "(1)" or "(2)" and solve for d"#
#(1)to35+3d=29#
#"subtract 35 from both sides"#
#rArr3d=29-35=-6rArrd=-2#
#"the solution is "(c,d)to(5,-2)#
#(B)" uses same approach as in "(A)#
#2x+3y=11to(1)#
#3x-5y=-12to(2)#
#"multiply "(1)" by 5 and "(2)" by3"#
#rArr10x+15y=55to(3)#
#9x-15y=-36to(4)#
#"add "(3)" and "(4)" term by term"#
#rArr19x=19rArrx=1#
#"substitute this value into "(1)#
#(1)to2+3y=11#
#rArr3y=11-2=9rArry=3#
#"the solution is "(x,y)to(1,3)#
