Question 1105396
<br>
As the other tutor showed, the system of equation we have for this problem is
(1) {{{4x+3y+2z = 15.50}}}
(2) {{{2x+6z = 28.60}}}
(3) {{{3x+4y = 6.80}}}<br>
When a system of 2 or more equations has all the equations in this form, I find it is nearly always less work to solve the system using elimination instead of substitution.  So here is what I would do with this problem....<br>
Variable x appears in all three equations.  So eliminate x between (1) and (2) and again between (1) and (3) to obtain two equations in y and z:<br>
Double equation (2) and subtract equation (1):
{{{4x+12z = 57.20}}}
{{{4x+3y+2z = 15.50}}}
(4) {{{-3y+10z = 41.70}}}<br>
Multiply (1) times 3 and (3) times 4 and subtract:
{{{12x+9y+6z = 46.50}}}
{{{12x+16y = 27.20}}}
(5) {{{-7y+6z = 19.30}}}<br>
Now eliminate z between equations (4) and (5): multiply (4) times 3 and (5) times -5 and add:
{{{-9y+30z = 125.10}}}
{{{35y-30z = -96.50}}}
{{{26y = 28.60}}}
(6) {{{y = 1.10}}}<br>
Now substitute (6) in (5) to solve for z:
{{{-7.70+6z = 19.30}}}
{{{6z = 27}}}
(7) {{{z = 4.50}}}<br>
And finally substitute (7) in (2) to solve for x:
{{{2x+27 = 28.60}}}
{{{2x = 1.60}}}
(8) {{{x = 0.80}}}<br>
(6), (7), and (8) tell us that...
an orange costs 0.80;
an apple costs 1.10; and
a pear costs $4.50