Solve the following system of equations?
1.) x - y = 3 2x - y = -1 Express x from the first equation: x = 3 + y and substitute it into the second one and solve for y: 2(3 + y) - y = -1 6 + 2y - y = -1 6 + y = -1 y = -7 x = 3 + y (from the first equation) x = 3 - 7 x = - 4 _________ 2.) 2x + 4y = 10 3x - 2y = -9 Here we can use the same method we used in the first question, but we can also do the following: keep the first equation as it is, and multiply the second one (both sides) by 2 (so that there are 4y in the first equation and -4y in the second one): 2x + 4y = 10 6x - 4y = -18 Now add the both equations up: add the left hand-sides up, and add the right hand-sides up. The y's will cancel: 4y - 4y = 0! 2x + 6x + 4y - 4y = 10 - 18 8x = 8 x = 1 Now substitute x = 1 in the first equation and solve it for y: 2·1 + 4y = 10 4y = 8 y = 2 __________ Hope this helps!
Solve the following system of equations.?
2x+y=3 x=2y-1 Substitute x=2y-1 into 2x+y=3. 2x+y=3 2(2y-1)+y=3 4y-2+y=3 5y=3+2 5y=5 y=1 Therefore if y=1 then x=2(1)-1=1 So x=1 and y=1. ANSWER:{(x=1,y=1)} (Next Question) 3x+2y-5=0 x=y+10 Again, substitute x=y+10. 3x+2y-5=0 3(y+10)+2y=5 3y+30+2y=5 5y=5-30 5y=-25 y=-5 Therefore if y=-5 then x=-5+10=5. So x=5 and y=-5. ANSWER{(x=5,y=-5)}
Solve the following system of equations. 2x – 3y + z = –1 3x + 2y + 2z = –1 x – y – 3z = –4?
Standard form: 2x-3y+z=-1; 3x+2y+2z=-1; x-y-3z=-4; substitute/eliminate x = +3/2y-1/2z-1/2 substitute/eliminate y = -1/13z+1/13 substitute/eliminate z = +1 Solution: x = -1; y = 0; z = 1; (x, y, z) = (-1, 0, 1)
How do you solve the system of equations 2x+2y=4 and 5y-3x=6?
Let the equations be Equation A and Equation B respectively ie.,2x + 2y = 4 … A5y - 3x = 6 … BNow lets make at least one variable of the both equations equal, so for that, lets multiply Equation A by 2 and Equation B by 3 and lets them as Equation C and Equation D:(3) * ( 2x + 2y ) = (3) * (4)6x + 6y = 12 … C(2) * ( 5y - 3x ) = (2) * (6)10y - 6x = 12 … DNow let us solve the new equations by adding them:6x + 6y + 10y - 6x = 12 + 1216y = 24y = 24 / 16y = 3 / 2Now let put this value of “y” in ANY of the above equations formed ie., Equation A ,Equation B, Equation C, Equation D. But I am putting it in Equation A:2x + 2y = 42x + (2) * ( 3 / 2 ) = 42x + 3 = 42x = 4 - 32x = 1x = 1 / 2So, the values of “x” and “y” which satisfy the both equations are as under:x = 1 / 2 , y = 3 / 2
How do I solve the following system of equations 2x + 3y = 5 3x + 7y = 15?
1) 2x +3y = 52) 3x +7y = 15Multiply equation 1 by 3 and equation 2 by 23) 6x +9y = 154) 6x +14y = 30Subtract equation 3 from equation 45y = 15y = 3Substitute for y in equation 12x +3*3 = 52x = -4x = -2****************************x = -2 and y = 3check answer by substituting for x and y in equation 23*-2 +7*3 = 1515 =15answer is ok******************************
How many solutions does the following system have: 6x - 3y = 9 2x - y = 3?
6x - 3y = 9 2x - y = 3 multiply the second equation by 3 6x - 3y = 9 3(2x - y) = 3*3 6x - 3y = 9 6x - 3y = 9 The equations are the same, than the system is indeterminate, it has infinite solutions. Bye gilvi